Demo Test
Complex Numbers

Question 1 :

Let a,b be two real numbers such that ab < 0 . IF the complex number $\frac{1+ai}{b+i}$ is of unit modulus and a + ib lies on the circle |z - 1| = |2z| , then a possible value of $\frac{1+\left[a\right]}{4b}$ , where [t] is greatest integer function, is :
1) $\left(\frac{1+\sqrt{7}}{4}\right)$
2) 1/2
3) 0
4) -1
3
ans : 3
$\frac{1+ai}{b+i}$ = 1 ⇒ |1 + ai | = |b+ i|
⇒ a 2 + 1 = b 2 + 1
⇒ a = b
⇒ ab < 0
⇒ (a + i b) lies on |z-1| = |2z|
⇒ |a + i b - 1| = 2 |a+ ib|
⇒ 6a2 + 2 a - 1 = 0
⇒ a = $\frac{-1+\sqrt{7}}{6}$ and $\frac{1-\sqrt{7}}{6}$
⇒ [a] = 0
$\frac{1+\left[a\right]}{4b}$ = 0
⇒ similarly when
⇒ [a] = -1
$\frac{1+\left[a\right]}{4b}$ = 0 = 0
If the center and radius of the circle $|\frac{z-2}{z-3}|=2$ are respectively (P,Q) and R , then 3(P+Q+R) is equal to
1) 12
2) 10
3) 11
4) 9
1
solution: ans : 1
|z - 2| = 2 |z-3|
⇒ x2 y 2 - 4x +4 = 4x2 4y 2 - 24x +36
⇒ x2 y 2 - 20/3x + 32/3 = 0
⇒ (P,Q) = (10/3 , 0) and using radius formula from the circles, we can find R = 2/3 therefore 3(P+Q+R) = 12
For two non-zero complex numbers z1 and z2 , if Re(z1z2) and , then which of the following are possible? A. Im(z1) > 0 and Im(z2) > 0 B. Im(z1) < 0 and Im(z2) > 0 C. Im(z1) > 0 and Im(z2) < 0 D. Im(z1) < 0 and Im(z2 ) < 0 Choose the correct answer from the options given below:
1) A & B
2) A & C
3) B & D
4) B & C
4
sol: ans : 4 Let, z1 = x1 + i y1 and z2 = x2 + i y2
⇒ z1z2 = x1x2 - y1y2 + i(x1y2 + x1y2)
Given, Re(z1 + z2) = 0
⇒ x1 + x2 = 0 x1x2 - y1y2 = 0 since x1 + x2 = 0 y1y2 = - xx2 So, multiplication of imaginary part's of z1 and z2 is negative. It means sign of y1 and y2 are opposite of each other. ∴ B and C are correct.
Let z be a complex number such that $|\frac{z-2i}{z+i}|=2,z\ne -i$ . Then lies on the circle of radius 2 and centre
1) (0, -2)
2)(0,0)
3) (0,2)
4) (2,0)
1
sol: ans:1
$|\frac{z-2i}{z+i}|=2$

$⇒\left(z-2i\right)\left(\overline{z}+2i\right)=4\left(z+i\right)\left(\overline{z}-i\right)$

$⇒2\overline{z}+2iz-2i\overline{z}+4=4\left(z\overline{z}-zi+\stackrel{―}{zi}+1\right)$

$⇒3z\overline{z}-6iz+6i\overline{z}=0$

$⇒2\overline{z}-2iz+2i\overline{z}=0$

$\therefore$ center $\left(-2i\right)$ or $\left(0,-2\right)$
Let z1 = 2 + 3 and z 2 = 3 + 4i . The set $S=\left\{z\in C:{|z-{z}_{1}|}^{2}-{|z-{z}_{2}|}^{2}={|{z}_{1}-{z}_{2}|}^{2}\right\}$ represents a
1) hyperbola with eccentricity = 2
2) straight line with the sum of its intercepts on the coordinates axes is -18
3) hyperbo;a with the length of transverse axis = 7
4) straight line with the sum of its intecepts on the coordinated axes = 14
4
sol: ans : 4
${|z-{z}_{1}|}^{2}-{|z-{z}_{2}|}^{2}={|{z}_{1}-{z}_{2}|}^{2}$

$⇒\left(x-2{\right)}^{2}+\left(y-3{\right)}^{2}-\left(x-3{\right)}^{2}-\left(y-4{\right)}^{2}=1+1$

$⇒-4x+4+9-6y-9+6x-16+8y=2$

$⇒2x+2y=14$

$⇒x+y=7$

The value of ${\left(\frac{1+\mathrm{sin}\frac{2\pi }{9}+i\mathrm{cos}\frac{2\pi }{9}}{1+\mathrm{sin}\frac{2\pi }{9}-i\mathrm{cos}\frac{2\pi }{9}}\right)}^{3}$ is:
1) $-\frac{1}{2}\left(1-i\sqrt{3}\right)$
2) $-\frac{1}{2}\left(\sqrt{3}-i\right)$
3) $\frac{1}{2}\left(1-i\sqrt{3}\right)$
4) $\frac{1}{2}\left(\sqrt{3}+i\right)$
2
solution: ans: 2
$z={\left(\frac{1+\mathrm{sin}\frac{2\pi }{9}+i\mathrm{cos}\frac{2\pi }{9}}{1+\mathrm{sin}\frac{2\pi }{9}-i\mathrm{cos}\frac{2\pi }{9}}\right)}^{3}$

$\begin{array}{cc}& 1+\mathrm{sin}\frac{2\pi }{9}+i\mathrm{cos}\frac{2\pi }{9}=1+\mathrm{cos}\frac{5\pi }{18}+i\mathrm{sin}\frac{5\pi }{18}\\ \\ & =1+2{\mathrm{cos}}^{2}\frac{5\pi }{36}-1+2i\mathrm{sin}\frac{5\pi }{36}\mathrm{cos}\frac{5\pi }{36}\\ \\ & =2\mathrm{cos}\frac{5\pi }{36}\left(\mathrm{cos}\frac{5\pi }{36}+i\mathrm{sin}\frac{5\pi }{36}\right)=2\mathrm{cos}\frac{5\pi }{36}{e}^{i\frac{5\pi }{36}}\\ \\ & z=-\frac{\sqrt{3}}{2}+\frac{1}{2}i=\frac{1}{2}\left(i-\sqrt{3}\right)=-\frac{1}{2}\left(\sqrt{3}-i\right)\end{array}$

Let $p,q\in R$ and ${\left(1-\sqrt{3}i\right)}^{200}={2}^{199}\left(p+iq\right),i=\sqrt{-1}$ then $p+q+{q}^{2}$ and $p-q+{q}^{2}$ are roots of the equation.
1) ${x}^{2}+4x-1=0$
2) ${x}^{2}-4x+1=0$
3) ${x}^{2}+4x+1=0$
4) ${x}^{2}-4x-1=0$
2
sol: ans:2 ${\left(1-\sqrt{3}i\right)}^{200}$

$={\left[2\left(\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)\right]}^{200}$

$={2}^{200}{\left(\mathrm{cos}\frac{\pi }{3}-i\mathrm{sin}\frac{\pi }{3}\right)}^{200}$

$={2}^{200}\left(\mathrm{cos}\frac{200\pi }{3}-i\mathrm{sin}\frac{200\pi }{3}\right)$

$={2}^{200}\left(\mathrm{cos}\left(66\pi +\frac{2\pi }{3}\right)-i\mathrm{sin}\left(66\theta +\frac{2\pi }{3}\right)\right)$

$={2}^{200}\left(\mathrm{cos}\frac{2\pi }{3}-i\mathrm{sin}\frac{2\pi }{3}\right)$

$={2}^{200}\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)$

$={2}^{199}\left(-1-\sqrt{3}i\right)$

$={2}^{199}\left(p+iq\right)$

$\therefore$ p = -1 and q = $-\sqrt{3}$

$p-q+{q}^{2}=-1+\sqrt{3}+3=2+\sqrt{3}=\alpha$

$p+q+{q}^{2}=-1-\sqrt{3}+3=2-\sqrt{3}=\beta$

$\therefore$ $\alpha +\beta =4$

$\alpha \beta =\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=4-3=1$

$\therefore$ ${x}^{2}-\left(\alpha +\beta \right)x+\alpha \beta =0$

$⇒{x}^{2}-4x+1=0$

If $z\ne 0$ be a complex number such that $|z-\frac{1}{z}|=2$ , then the maximum value of $|z|$ is :
1) $\sqrt{2}$
2) 1
3) $\sqrt{2}-1$
4) $\sqrt{2}+1$
4
solution: ans: 4

$||{z}_{1}|-|{z}_{2}||\le |{z}_{1}+{z}_{2}|\le |{z}_{1}|+|{z}_{2}|$

$\therefore$ $||z|-\frac{1}{|z|}|\le |z-\frac{1}{z}|$

$⇒||z|-\frac{1}{|z|}|\le 2$

$⇒|\frac{|z{|}^{2}-1}{|z|}|\le 2$

$⇒-2\le \frac{|z{|}^{2}-1}{|z|}\le 2$

$\therefore$ $\frac{|z{|}^{2}-1}{|z|}\le 2$

$⇒|z{|}^{2}-1\le 2|z|$

$⇒|z{|}^{2}-2|z|-1\le 0$

$⇒|z{|}^{2}-2|z|+1-2\le 0$

$⇒\left(|z|-1{\right)}^{2}-2\le 0$

$⇒-\sqrt{2}\le |z|-1\le \sqrt{2}$

$⇒1-\sqrt{2}\le |z|\le 1+\sqrt{2}$ ------(1)

or

$-2\le \frac{|z{|}^{2}-1}{|z|}$

$⇒|z{|}^{2}-1\le -2|z|$

$⇒|z{|}^{2}+2|z|-1\le 0$

$⇒|z{|}^{2}+2|z|+1-2\le 0$

$⇒\left(|z|+1{\right)}^{2}-2\le 0$

$⇒-\sqrt{2}\le |z|+1\le +\phantom{\rule{0.167em}{0ex}}\sqrt{2}$

$⇒-\sqrt{2}-1\le |z|\le \sqrt{2}-1$ ------ (2) From (1) and (2) we get, Maximum value of $|z|=\sqrt{2}+1$ and minimum value of $|z|=-\sqrt{2}-1$
Let $S=\left\{z=x+iy:|z-1+i|\ge |z|,|z|<2,|z+i|=|z-1|\right\}$ . Then the set of all values of $x$ , for which $w=2x+iy\in S$ for some $y\in R$ , is
1) $\left(-\sqrt{2},\frac{1}{2\sqrt{2}}\right]$
2) $\left(-\frac{1}{\sqrt{2}},\frac{1}{4}\right]$
3) $\left(-\sqrt{2},\frac{1}{2}\right]$
4) $\left(-\frac{1}{\sqrt{2}},\frac{1}{2\sqrt{2}}\right]$
2
If $z=2+3i$ , then ${z}^{5}+\left(\overline{z}{\right)}^{5}$ is equal to :
1) 244
2) 224
3) 245
4) 265
1
ans: 1
$z=\left(2+3i\right)$

$⇒{z}^{5}=\left(2+3i\right){\left({\left(2+3i\right)}^{2}\right)}^{2}$

$=\left(2+3i\right)\left(-5+12i{\right)}^{2}$

$=\left(2+3i\right)\left(-119-120i\right)$

$=-238-240i-357i+360$

$=122-597i$

${\stackrel{―}{z}}^{5}=122+597i$

${z}^{5}+{\stackrel{―}{z}}^{5}=244$

Let S be the set of all $\left(\alpha ,\beta \right),\pi <\alpha ,\beta <2\pi$ , for which the complex number $\frac{1-i\mathrm{sin}\alpha }{1+2i\mathrm{sin}\alpha }$ is purely imaginary and $\frac{1+i\mathrm{cos}\beta }{1-2i\mathrm{cos}\beta }$ is purely real. Let ${Z}_{\alpha \beta }=\mathrm{sin}2\alpha +i\mathrm{cos}2\beta ,\left(\alpha ,\beta \right)\in S$ . Then $\sum _{\left(\alpha ,\beta \right)\in S}\left(i{Z}_{\alpha \beta }+\frac{1}{i{\overline{Z}}_{\alpha \beta }}\right)$ is equal to :
1) 3
2) 3i
3) 1
4) 2 - i
3
solution: ans:3
$\because$ $\frac{1-i\mathrm{sin}\alpha }{1+2i\mathrm{sin}\alpha }$ is purely imaginary

$\frac{1-i\mathrm{sin}\alpha }{1+2i\mathrm{sin}\alpha }+\frac{1+i\mathrm{sin}\alpha }{1-2i\mathrm{sin}\alpha }=0$

$⇒1-2{\mathrm{sin}}^{2}\alpha =0$

$\therefore$ $\alpha =\frac{5\pi }{4},\phantom{\rule{0.167em}{0ex}}\frac{7\pi }{4}$

$\frac{1+i\mathrm{cos}\beta }{1-2i\mathrm{cos}\beta }$

$\frac{1+i\mathrm{cos}\beta }{1-2i\mathrm{cos}\beta }-\frac{1-i\mathrm{cos}\beta }{1+2i\mathrm{cos}\beta }=0$

$⇒\mathrm{cos}\beta =0$

$\therefore$ $\beta =\frac{3\pi }{2}$

$\therefore$ $S=\left\{\left(\frac{5\pi }{2},\frac{3\pi }{2}\right),\left(\frac{7\pi }{4},\frac{3\pi }{2}\right)\right\}$

${Z}_{\alpha \beta }=1-i$

${Z}_{\alpha \beta }=-1-i$

$\therefore$ $\sum _{\left(\alpha ,\beta \right)\in S}\left(i{Z}_{\alpha \beta }+\frac{1}{i{\stackrel{―}{Z}}_{\alpha \beta }}\right)=i\left(-2i\right)+\frac{1}{i}\left[\frac{1}{1+i}+\frac{1}{-1+i}\right]$

$=2+\frac{1}{i}\frac{2i}{-2}=1$
Let the minimum value ${v}_{0}$ of $v=|z{|}^{2}+|z-3{|}^{2}+|z-6i{|}^{2},z\in \mathbb{C}$ is attained at $z={z}_{0}$ . Then ${|2{z}_{0}^{2}-{\overline{z}}_{0}^{3}+3|}^{2}+{v}_{0}^{2}$ is equal to
1) 1000
2) 1024
3) 1105
1196
1
solution: ans: 1
$z=x+iy$

$v={x}^{2}+{y}^{2}+\left(x-3{\right)}^{2}+{y}^{2}+{x}^{2}+\left(y-6{\right)}^{2}$

$=\left(3{x}^{2}-6x+9\right)+\left(3{y}^{2}-12y+36\right)$

$=3\left({x}^{2}+{y}^{2}-2x-4y+15\right)$

$=3\left[\left(x-1{\right)}^{2}+\left(y-2{\right)}^{2}+10\right]$

${v}_{min}$

$z=1+2i={z}_{0}$

${v}_{0}=30$

$|2\left(1+2i{\right)}^{2}-\left(1-2i{\right)}^{3}+3{|}^{2}+900$

$=|2\left(-3+4i\right)-\left(1-8{i}^{3}-6i\left(1-2i\right)+3{|}^{2}+900$

$=|-6+8i-\left(1+8i-6i-12\right)+3{|}^{2}+900$

$=|8+6i{|}^{2}+900$

$=1000$
If $z=x+iy$ satisfies $|z|-2=0$ and $|z-i|-|z+5i|=0$ , then
1) $x+2y-4=0$
2) ${x}^{2}+y-4=0$
3) $x+2y+4=0$
4) ${x}^{2}-y+3=0$
3
sol: ans:3
$|z-i|=|z+5i|$

So, $z$ lies on ${\perp }^{}$ bisector of $\left(0,1\right)$ and $\left(0,-5\right)$ i.e., line $y=-2$

$|z|=2$

$⇒z=-2i$

$x=0$

$y=-2$

$x+2y+4=0$
If $\alpha ,\beta ,\gamma ,\delta$ are the roots of the equation ${x}^{4}+{x}^{3}+{x}^{2}+x+1=0$ , then ${\alpha }^{2021}+{\beta }^{2021}+{\gamma }^{2021}+{\delta }^{2021}$ is equal to :
1) -4
2) -1
3) 1
4) 4
2
solution: ans= 2
${x}^{5}=1$

${x}^{5}-1=0$

$⇒\left(x-1\right)\left({x}^{4}+{x}^{3}+{x}^{2}+x+1\right)=0$

${x}^{4}+{x}^{3}+{x}^{2}+x+1=0$ has roots $\alpha$ , $\beta$ , $\gamma$ and 8.
$\therefore$ Roots of ${x}^{5}-1=0$ are 1, $\alpha$ , $\beta$ , $\gamma$ , 8.We know, Sum of pth power of nth roots of unity = 0. (If p is not multiple of n) or n (If p is multiple of n)
$\therefore$ Here, Sum of pth power of nth roots of unity $={1}^{p}+{\alpha }^{p}+{\beta }^{p}+{\gamma }^{p}+{8}^{p}=\left\{\begin{array}{ccc}0& ;& \mathrm{If}\phantom{\rule{0.167em}{0ex}}p\phantom{\rule{0.167em}{0ex}}\mathrm{is}\phantom{\rule{0.167em}{0ex}}\mathrm{not}\phantom{\rule{0.167em}{0ex}}\mathrm{multiple}\phantom{\rule{0.167em}{0ex}}\mathrm{of}\phantom{\rule{0.167em}{0ex}}5\\ 5& ;& \mathrm{If}\phantom{\rule{0.167em}{0ex}}p\phantom{\rule{0.167em}{0ex}}\mathrm{is}\phantom{\rule{0.167em}{0ex}}\mathrm{multiple}\phantom{\rule{0.167em}{0ex}}\mathrm{of}\phantom{\rule{0.167em}{0ex}}5\end{array}$

$p=2021$ is not a multiple of 5.

$\therefore$ ${1}^{2021}+{\alpha }^{2021}+{\beta }^{2021}+{\gamma }^{2021}+{8}^{2021}=0$

$⇒{\alpha }^{2021}+{\beta }^{2021}+{\gamma }^{2021}+{8}^{2021}=-1$
For $n\in N$ , let ${S}_{n}=\left\{z\in C:|z-3+2i|=\frac{n}{4}\right\}$ and ${T}_{n}=\left\{z\in C:|z-2+3i|=\frac{1}{n}\right\}$ . Then the number of elements in the set $\left\{n\in N:{S}_{n}\cap {T}_{n}=\varphi \right\}$ is :
1) 0
2) 2
3) 3
4) 4
4
solution: ans: 4
${S}_{n}=\left\{z\in C:|z-3+2i|=\frac{n}{4}\right\}$ represents a circle with centre ${C}_{1}\left(3,-2\right)$ and radius ${r}_{1}=\frac{n}{4}$

similarly ${T}_{n}$ represents circle with centre ${C}_{2}\left(2,-3\right)$ and radius ${r}_{2}=\frac{1}{n}$

The real part of the complex number $\frac{{\left(1+2i\right)}^{8}.{\left(1-2i\right)}^{2}}{\left(3+2i\right).\stackrel{―}{\left(4-6i\right)}}$ is equal to :

1) $\frac{500}{13}$
2) $\frac{110}{13}$
3) $\frac{55}{6}$
4) $\frac{550}{13}$
4
solution: ans:4
$\frac{{\left(1+2i\right)}^{8}.{\left(1-2i\right)}^{2}}{\left(3+2i\right).\stackrel{―}{\left(4-6i\right)}}$

$=\frac{{\left(1+2i\right)}^{2}{\left(1-2i\right)}^{2}{\left(1+2i\right)}^{6}}{\left(3+2i\right)\left(4+6i\right)}$

$=\frac{{\left(1-4{i}^{2}\right)}^{2}{\left(1+2i\right)}^{6}}{12+18i+8i+12{i}^{2}}$

$=\frac{{\left(1+5\right)}^{2}{\left[{\left(1+2i\right)}^{2}\right]}^{3}}{12+26i-12}$

$=\frac{25{\left(1+4{i}^{2}+4i\right)}^{3}}{26i}$

$=\frac{25{\left(1-4+4i\right)}^{3}}{26i}$

$=\frac{25{\left(-3+4i\right)}^{3}}{26i}$

$=\frac{25}{26i}\left[{\left(-3\right)}^{3}+{\left(4i\right)}^{3}+3.{\left(-3\right)}^{2}.4i+3\left(-3\right).{\left(4i\right)}^{2}\right]$

$=\frac{25}{26i}\left(-27-64i+108i+144\right)$

$=\frac{25}{26i}\left(117+44i\right)$

$=\frac{25i}{26{i}^{2}}\left(117+44i\right)$

$=\frac{25i}{-26}\left(117+44i\right)$

$=\frac{25×117i}{-26}-\frac{25×44{i}^{2}}{26}$

$=\frac{25×117i}{-26}+\frac{22×25}{13}$

$=\frac{25×117i}{-26}+\frac{550}{13}$

$\therefore$ real part $=\frac{550}{13}$

Let $\alpha$ and $\beta$ be the roots of the equation x2 + (2i - 1) = 0. Then, the value of |$\alpha$ 8 + $\beta$ 8| is equal to:
1) 50
2) 250
3) 1250
4) 1500
1
solution: ans: 1
${x}^{2}+\left(2i-1\right)=0$

$⇒{x}^{2}=1-2i$

as $\alpha$ and $\beta$ are roots of the equation

${\alpha }^{2}=1-2i$

${\beta }^{2}=1-2i$

$\therefore$ ${\alpha }^{2}={\beta }^{2}=1-2i$

$\therefore$ $|{\alpha }^{2}|=\sqrt{{1}^{2}+{\left(-2\right)}^{2}}=\sqrt{15}$

$|{\alpha }^{8}+{\beta }^{8}|$

$|{\alpha }^{8}+{\alpha }^{8}|$

$=2|{\alpha }^{8}|$

$=2|{\alpha }^{2}{|}^{4}$

$=2{\left(\sqrt{5}\right)}^{4}$

$=2×25$

$=50$
The area of the polygon, whose vertices are the non-real roots of the equation $\stackrel{―}{z}=i{z}^{2}$ is :
1) $\frac{3\sqrt{3}}{4}$
2) $\frac{3\sqrt{3}}{2}$
3) $\frac{3}{2}$
4) $\frac{3}{4}$
1
sol: ans: 1
$\stackrel{―}{z}=i{z}^{2}$

$z=x+iy$

$x-iy=i\left({x}^{2}-{y}^{2}+2xiy\right)$

$x-iy=i\left({x}^{2}-{y}^{2}\right)-2xy$

$\therefore$ $x=-2yx$

${x}^{2}-{y}^{2}=-y$

case - 1
$x=0$

$y=-\frac{1}{2}$

$x=0$

$-{y}^{2}=-y$

$y=0,1$

$y=-\frac{1}{2}$
case - 2
$⇒{x}^{2}-\frac{1}{4}=\frac{1}{2}⇒x=±\frac{\sqrt{3}}{2}$

$x=\left\{0,i,\frac{\sqrt{3}}{2}-\frac{i}{2},\frac{-\sqrt{3}}{2}-\frac{i}{2}\right\}$

area of polygon = $=\frac{1}{2}|\begin{array}{ccc}0& 1& 1\\ \frac{\sqrt{3}}{2}& \frac{-1}{2}& 1\\ \frac{-\sqrt{3}}{2}& \frac{-1}{2}& 1\end{array}|$

$=\frac{1}{2}|-\sqrt{3}-\frac{\sqrt{3}}{2}|=\frac{3\sqrt{3}}{4}$
Let z1 and z2 be two complex numbers such that ${\stackrel{―}{z}}_{1}=i{\stackrel{―}{z}}_{2}$ and $\mathrm{arg}\left(\frac{{z}_{1}}{{\stackrel{―}{z}}_{2}}\right)=\pi$ . Then
1) $\mathrm{arg}{z}_{2}=\frac{\pi }{4}$
2) $\mathrm{arg}{z}_{2}=-\frac{3\pi }{4}$
3) $\mathrm{arg}{z}_{1}=\frac{\pi }{4}$
4) $\mathrm{arg}{z}_{1}=-\frac{3\pi }{4}$
3
solution: ans:3
$\frac{{z}_{1}}{{z}_{2}}=-i⇒{z}_{1}=-i{z}_{2}$

$⇒\mathrm{arg}\left({z}_{1}\right)=-\frac{\pi }{2}+\mathrm{arg}\left({z}_{2}\right)$

$\mathrm{arg}\left({z}_{1}\right)-\mathrm{arg}\left({\stackrel{―}{z}}_{2}\right)=\pi$

$⇒\mathrm{arg}\left({z}_{1}\right)+\mathrm{arg}\left({z}_{2}\right)=\pi$

By solveing above equations we get, $\mathrm{arg}\left({z}_{1}\right)=\frac{\pi }{4}$ and $\mathrm{arg}\left({z}_{2}\right)=\frac{3\pi }{4}$
Let a circle C in complex plane pass through the points ${z}_{1}=3+4i$ , ${z}_{2}=4+3i$ and ${z}_{3}=5i$ . If $z\left(\ne {z}_{1}\right)$ is a point on C such that the line through z and z1 is perpendicular to the line through z2 and z3, then arg(z) is equal to:
1) ${\mathrm{tan}}^{-1}\left(\frac{2}{\sqrt{5}}\right)-\pi$
2) ${\mathrm{tan}}^{-1}\left(\frac{24}{7}\right)-\pi$
3) ${\mathrm{tan}}^{-1}\left(3\right)-\pi$
4) ${\mathrm{tan}}^{-1}\left(\frac{3}{4}\right)-\pi$
2
solution: ans: 2
${z}_{1}=3+4i$

${z}_{2}=4+3i$

${z}_{3}=5i$

$C\equiv {x}^{2}+{y}^{2}=25$

Let z(x,y)
$z\left(x,y\right)$

$⇒\left(\frac{y-4}{x-3}\right)\left(\frac{2}{-4}\right)=-1$

$⇒y=2x-2\equiv L$

$\therefore$ z is intersection of C and L

$⇒z\equiv \left(\frac{-7}{5},\frac{-24}{5}\right)$

$\therefore$ $Arg\left(z\right)=-\pi +{\mathrm{tan}}^{-1}\left(\frac{24}{7}\right)$
If ${\left(\sqrt{3}+i\right)}^{100}={2}^{99}\left(p+iq\right)$ , then p and q are roots of the equation :
1) ${x}^{2}-\left(\sqrt{3}-1\right)x-\sqrt{3}=0$
2) ${x}^{2}+\left(\sqrt{3}+1\right)x+\sqrt{3}=0$
3) ${x}^{2}+\left(\sqrt{3}-1\right)x-\sqrt{3}=0$
4) ${x}^{2}-\left(\sqrt{3}+1\right)x+\sqrt{3}=0$
1
solution: ans:1 ${\left(2{e}^{i\pi /6}\right)}^{100}={2}^{99}\left(p+iq\right)$

${2}^{100}\left(\mathrm{cos}\frac{50\pi }{3}+i\mathrm{sin}\frac{50\pi }{3}\right)={2}^{99}\left(p+iq\right)$

$p+iq=2\left(\mathrm{cos}\frac{2\pi }{3}+i\mathrm{sin}\frac{2\pi }{3}\right)$

p = $-1$

q = $\sqrt{3}$

${x}^{2}-\left(\sqrt{3}-1\right)x-\sqrt{3}=0$

If z and $\omega$ are two complex numbers such that $|z\omega |=1$ and $\mathrm{arg}\left(z\right)-\mathrm{arg}\left(\omega \right)=\frac{3\pi }{2}$ , then $\mathrm{arg}\left(\frac{1-2\stackrel{―}{z}\omega }{1+3\stackrel{―}{z}\omega }\right)$ is : (Here arg(z) denotes the principal argument of complex number z)
1) $\mathrm{arg}\left(\frac{1-2\stackrel{―}{z}\omega }{1+3\stackrel{―}{z}\omega }\right)$
2) $\frac{\pi }{4}$
3) $-\frac{3\pi }{4}$
4) $-\frac{\pi }{4}$
2
solution: ans: 2
$\frac{3\pi }{4}$

$|z\omega |=1$

$⇒$

$|z|=r$

$|\omega |=\frac{1}{r}$

let $\mathrm{arg}\left(z\right)=\theta$

$\therefore$ $\mathrm{arg}\left(\omega \right)=\left(\theta -\frac{3\pi }{2}\right)$

$z=r{e}^{i\theta }$

$⇒\stackrel{―}{z}=r{e}^{i\theta }$

$\omega =\frac{1}{r}{e}^{i\left(\theta -\frac{3\pi }{2}\right)}$

now consider $\frac{1-2\stackrel{―}{z}\omega }{1+3\stackrel{―}{z}\omega }=\frac{1-2{e}^{i\left(-\frac{3\pi }{2}\right)}}{1+3{e}^{i\left(-\frac{3\pi }{2}\right)}}=\left(\frac{1-2i}{1+3i}\right)$

$=\frac{\left(1-2i\right)\left(1-3i\right)}{\left(1+3i\right)\left(1-3i\right)}=-\frac{1}{2}\left(1+i\right)$

$\therefore$ $prin\mathrm{arg}\left(\frac{1-2\stackrel{―}{z}\omega }{1+3\stackrel{―}{z}\omega }\right)$

$=prin\mathrm{arg}\left(\frac{1-2\stackrel{―}{z}\omega }{1+3\stackrel{―}{z}\omega }\right)$

$=\left(-\frac{1}{2}\left(1+i\right)\right)$

$=-\left(\pi -\frac{\pi }{4}\right)=\frac{-3\pi }{4}$

The least value of |z| where z is complex number which satisfies the inequality $\mathrm{exp}\left(\frac{\left(|z|+3\right)\left(|z|-1\right)}{||z|+1|}{\mathrm{log}}_{e}2\right)\ge {\mathrm{log}}_{\sqrt{2}}|5\sqrt{7}+9i|,i=\sqrt{-1}$ , is equal to :
1) 8
2) 3
3) 2
4) $\sqrt{5}$
2
solution: ans: 2
Let |z| = t and t $\ge$ 0

${e}^{\frac{\left(t+3\right)\left(t-1\right)}{t+1}{\mathrm{log}}_{e}2}\ge {\mathrm{log}}_{\sqrt{2}}16=8$ $\because$t+1 $>$ 0

${2}^{\frac{\left(t+3\right)\left(t-1\right)}{t+1}}\ge {2}^{3}$

$\frac{\left(t+3\right)\left(t-1\right)}{t+1}\ge 3$

${t}^{2}+2t-3\ge 3t+3$

${t}^{2}-t-6\ge 0$

$t\in \left(-\mathrm{\infty },-2\right)\cup \left[3,\mathrm{\infty }\right)$

$\ge$

$\therefore$ $t\in \left[3,\mathrm{\infty }\right)$
Let a complex number z, |z| $\ne$ 1, satisfy ${\mathrm{log}}_{\frac{1}{\sqrt{2}}}\left(\frac{|z|+11}{{\left(|z|-1\right)}^{2}}\right)\le 2$ . Then, the largest value of |z| is equal to ____________.
1) 5
2) 8
3) 6
4) 7
4
solution: ans: 4 ${\mathrm{log}}_{\frac{1}{\sqrt{2}}}\left(\frac{|z|+11}{{\left(|z|-1\right)}^{2}}\right)\le 2$

$\frac{|z|+11}{{\left(|z|-1\right)}^{2}}\ge \frac{1}{2}$

$2|z|+22\ge \left(|z|-1{\right)}^{2}$

$2|z|+22\ge \phantom{\rule{0.167em}{0ex}}|z{|}^{2}-2|z|+1$

$|z{|}^{2}-4|z|-21\le 0$

$\left(|z|-7\right)\left(|z|+3\right)\le 0$

$⇒\phantom{\rule{0.167em}{0ex}}|z|\le 7$

$\therefore$

$|z{|}_{max}=7$
If $\alpha$ , $\beta$$\in$ R are such that 1 - 2i (here i2 = - 1) is a root of z2 + $\alpha$ z + $\beta$ = 0, then ($\alpha$ - $\beta$ ) is equal to :
1) -7
2) 7
3) 3
4) -3
1
solution: ans: 1
1 - 2i is the root of the equation. So other root is 1 + 2i $\therefore$ Sum of roots = 1 - 2i + 1 + 2i = 2 = - $\alpha$ Product of roots = (1 - 2i)(1 + 2i) = 1 - 4i2 = 5 = $\beta$ $\therefore$ $\alpha$ - $\beta$ = -7
Let the lines (2 - i)z = (2 + i)$\stackrel{―}{z}$ and (2 + i)z + (i - 2)$\stackrel{―}{z}$ - 4i = 0, (here i2 = - 1) be normal to a circle C. If the line iz + $\stackrel{―}{z}$ + 1 + i = 0 is tangent to this circle C, then its radius is :
1) $\frac{3}{2\sqrt{2}}$
2) $3\sqrt{2}$
3) $\frac{1}{2\sqrt{2}}$
4) $\frac{3}{\sqrt{2}}$
1
solution: ans:1
$\left(2-i\right)z=\left(2+i\right)\stackrel{―}{z}$

$⇒\left(2-i\right)\left(x+iy\right)=\left(2+i\right)\left(x-iy\right)$

$⇒2x-ix+2iy+y=2x+ix-2-iy+y$

$⇒2ix-4iy=0$

${L}_{1}:x-2y=0$

$⇒\left(2+i\right)z+\left(i-2\right)\stackrel{―}{z}-4i=0$

$⇒\left(2+i\right)\left(x+iy\right)+\left(i-2\right)\left(x-iy\right)-4i=0$

$⇒2x+ix+2iy-y+ix-2x+y+2iy-4i=0$

$⇒2ix+4iy-4i=0$

${L}_{2}:x+2y-2=0$

Solve L1 and L2: x = 1, $4y=2,y=\frac{1}{2}$

$\therefore$ $x=1$

centre $\left(1,\frac{1}{2}\right)$

${L}_{3}:iz+\stackrel{―}{z}+1+i=0$

$⇒i\left(x+iy\right)+x-iy+1+i=0$

$⇒ix-y+x-iy+1+i=0$

$⇒\left(x-y+1\right)+i\left(x-y+1\right)=0$

Radius = distance from $\left(1,\frac{1}{2}\right)$ to $x-y+1=0$

$⇒$

$r=\frac{1-\frac{1}{2}+1}{\sqrt{2}}$

$⇒$

$r=\frac{3}{2\sqrt{2}}$
The value of ${\left(\frac{-1+i\sqrt{3}}{1-i}\right)}^{30}$ is :
- 215i
- 215
215i
65
1
solution: ans:1

${\left(\frac{-1+i\sqrt{3}}{1-i}\right)}^{30}$

=${\left(\frac{2\omega }{1-i}\right)}^{30}$

=$\frac{{2}^{30}.{\omega }^{30}}{{\left({\left(1-i\right)}^{2}\right)}^{15}}$

=$\frac{{2}^{30}.1}{{\left(1+{i}^{{}^{2}}-2i\right)}^{15}}$

=$\frac{{2}^{30}.1}{-{2}^{15}.{i}^{15}}$

= - 215i
If a and b are real numbers such that ${\left(2+\alpha \right)}^{4}=a+b\alpha$ where $\alpha =\frac{-1+i\sqrt{3}}{2}$ then a + b is equal to:
1) 33
2) 9
3) 24
4) 57
2
solution: ans: 2
$\alpha =\omega$

given $\alpha =\frac{-1+i\sqrt{3}}{2}$

$⇒\left(2+\omega {\right)}^{4}=a+b\omega \phantom{\rule{0.167em}{0ex}}\left({\omega }^{3}=1\right)$

$⇒{2}^{4}+{4.2}^{3}\omega +{6.2}^{2}{\omega }^{3}+4.2.\phantom{\rule{0.167em}{0ex}}{\omega }^{3}+{\omega }^{4}=a+b\omega$

$⇒16+32\omega +24{\omega }^{2}+8+\omega =a+b\omega$

$⇒24+24{\omega }^{2}+33\omega =a+b\omega$

$⇒-24\omega +33\omega =a+b\omega$

$⇒a=0,\phantom{\rule{0.167em}{0ex}}b=9$
Let $u=\frac{2z+i}{z-ki}$ , z = x + iy and k > 0. If the curve represented by Re(u) + Im(u) = 1 intersects the y-axis at the points P and Q where PQ = 5, then the value of k is :
2
4
1/2
3/2
1
solution: ans:1 given z =x + iy

$u=\frac{2z+i}{z-ki}$

$=\frac{2\left(x+iy\right)+i}{\left(x+iy\right)-ki}$

$=\frac{2x+i\left(2y+1\right)}{x+i\left(y-k\right)}×\frac{x-i\left(y-k\right)}{x-i\left(y-k\right)}$

$=\frac{2{x}^{2}+\left(2y+1\right)\left(y-k\right)+i\left(2xy+x-2xy+2kx\right)}{{x}^{2}+{\left(y-k\right)}^{2}}$

Given, $\mathrm{Re}\left(u\right)+\mathrm{Im}\left(u\right)=1$

$⇒\frac{2{x}^{2}+\left(2y+1\right)\left(y-k\right)}{{x}^{2}+{\left(y-k\right)}^{2}}+\frac{x+2kx}{{x}^{2}+{\left(y-k\right)}^{2}}=1$

$⇒2{x}^{2}+\left(2y+1\right)\left(y-k\right)+x+2kx={x}^{2}+\left(y-k{\right)}^{2}$

This curve intersect the y-axis at point P and Q, so at point P and Q x = 0 Putting x = 0 at the above equation, $\therefore$ $\left(2y+1\right)\left(y-k\right)=\left(y-k{\right)}^{2}$

$⇒2{y}^{2}+y-2yk-k={y}^{2}+{k}^{2}-2ky$

$⇒{y}^{2}+y-\left(k+{k}^{2}\right)=0$

Let roots of this quadratic equation y1 and y2 $\therefore$ Point P (0, y1) and Q (0, y2) ${y}_{{}_{1}}+{y}_{2}=1$

${y}_{1}{y}_{2}=-k-{k}^{2}$

$\therefore$ $\left({y}_{1}-{y}_{2}{\right)}^{2}=\left({y}_{1}+{y}_{2}{\right)}^{2}-4{y}_{1}{y}_{2}$

$=1+4k+4{k}^{2}$

$⇒|{y}_{1}-{y}_{2}|=\sqrt{1+4k+4{k}^{2}}$

Given PQ = 5 $⇒|{y}_{1}-{y}_{2}|=5$

$⇒\sqrt{1+4k+4{k}^{2}}=5$

$⇒{k}^{2}+k-6=0$

$⇒k=-3,\phantom{\rule{0.167em}{0ex}}2$

give k is greater than 0 so, k= 2
If z1 , z2 are complex numbers such that Re(z1) = |z1 – 1|, Re(z2) = |z2 – 1| , and arg(z1 - z2) = $\frac{\pi }{6}$ , then Im(z1 + z2 ) is equal to :
1) $\frac{\sqrt{3}}{2}$
2) $\frac{1}{\sqrt{3}}$
3) $\frac{2}{\sqrt{3}}$
4) $2\sqrt{3}$
4
solution: ans: 4
${z}_{1}={x}_{1}+i{y}_{1},\phantom{\rule{0.167em}{0ex}}{z}_{2}={x}_{2}+i{y}_{2}$

Given Re(z1) = |z1 – 1| $\therefore$x1 = |(x1 - 1) + iy1|

$⇒$ x1 $\sqrt{{\left({x}_{1}-1\right)}^{2}+{y}_{1}^{2}}$

$⇒$

${{x}_{1}}^{2}=\left({x}_{1}-1{\right)}^{2}+{{y}_{1}}^{2}$

$⇒{{y}_{1}}^{2}-2{x}_{1}+1=0$

Also given Re(z2) = |z2 – 1| $\therefore$ x2 = |(x2 - 1) + iy2| $⇒$

Performing equation (1) - (2), ${{x}_{2}}^{2}=\left({x}^{2}-1{\right)}^{2}+{{y}_{2}}^{2}$

$⇒$

${y}_{2}^{2}-2{x}_{2}-1=0$

$\left({{y}_{1}}^{2}-{{y}_{2}}^{2}\right)+2\left({x}_{2}-{x}_{1}\right)=0$

$⇒$

$\left({y}_{1}+{y}_{2}\right)\left({y}_{1}-{y}_{2}\right)=2\left({x}_{1}-{x}_{2}\right)$

$⇒$

${y}_{1}+{y}_{2}=2\left(\frac{{x}_{1}-{x}_{2}}{{y}_{1}-{y}_{2}}\right)$

given, $\mathrm{arg}\left({z}_{1}-{z}_{2}\right)=\frac{\pi }{6}$

$⇒$

${\mathrm{tan}}^{-1}\left(\frac{{y}_{1}-{y}_{2}}{{x}_{1}-{x}_{2}}\right)=\frac{\pi }{6}$

$⇒\frac{{y}_{1}-{y}_{2}}{{x}_{1}-{x}_{2}}=\frac{1}{\sqrt{3}}$

$\therefore$ ${y}_{1}+{y}_{2}=2\sqrt{3}$
The imaginary part of ${\left(3+2\sqrt{-54}\right)}^{\frac{1}{2}}-{\left(3-2\sqrt{-54}\right)}^{\frac{1}{2}}$ can be :
1) -2$\sqrt{6}$
2) 6
3) $\sqrt{6}$
4) - $\sqrt{6}$
1
solution: ans:1

$3+2\sqrt{-54}$

$=9-6+2\sqrt{-54}$

$=9+{\left(\sqrt{6}i\right)}^{2}+2.3.\sqrt{6}i$

$={3}^{2}+{\left(\sqrt{6}i\right)}^{2}+2.3.\sqrt{6}i$

$={\left(3+\sqrt{6}i\right)}^{2}$

Similarly, $\left(3-2\sqrt{-54}\right)={\left(3-\sqrt{6}i\right)}^{2}$

$\therefore {\left(3+2\sqrt{-54}\right)}^{\frac{1}{2}}-{\left(3-2\sqrt{-54}\right)}^{\frac{1}{2}}$

$=±\left(3+\sqrt{6}i\right)-\left[±\left(3-\sqrt{6}i\right)\right]$

$=6,-6,2\sqrt{6}i,-2\sqrt{6}i$

$\therefore$ Possible imaginary parts are $2\sqrt{6}i,-2\sqrt{6}i$
The value of ${\left(\frac{1+\mathrm{sin}\frac{2\pi }{9}+i\mathrm{cos}\frac{2\pi }{9}}{1+\mathrm{sin}\frac{2\pi }{9}-i\mathrm{cos}\frac{2\pi }{9}}\right)}^{3}$ is :
1) $\frac{1}{2}\left(\sqrt{3}-i\right)$
2) -$\frac{1}{2}\left(\sqrt{3}-i\right)$
3) $-\frac{1}{2}\left(1-i\sqrt{3}\right)$
4) $\frac{1}{2}\left(1-i\sqrt{3}\right)$
2
solution: ans: 2
${\left(\frac{1+\mathrm{sin}\frac{2\pi }{9}+i\mathrm{cos}\frac{2\pi }{9}}{1+\mathrm{sin}\frac{2\pi }{9}-i\mathrm{cos}\frac{2\pi }{9}}\right)}^{3}$

=${\left(\frac{1+\mathrm{sin}\left(\frac{\pi }{2}-\frac{5\pi }{18}\right)+i\mathrm{cos}\left(\frac{\pi }{2}-\frac{5\pi }{18}\right)}{1+\mathrm{sin}\left(\frac{\pi }{2}-\frac{5\pi }{18}\right)-i\mathrm{cos}\left(\frac{\pi }{2}-\frac{5\pi }{18}\right)}\right)}^{3}$

=${\left(\frac{1+\mathrm{cos}\left(\frac{5\pi }{18}\right)+i\mathrm{sin}\left(\frac{5\pi }{18}\right)}{1+\mathrm{cos}\left(\frac{5\pi }{18}\right)-i\mathrm{sin}\left(\frac{5\pi }{18}\right)}\right)}^{3}$

=${\left(\frac{2{\mathrm{cos}}^{2}\left(\frac{5\pi }{36}\right)+2i\mathrm{sin}\left(\frac{5\pi }{36}\right)\mathrm{cos}\left(\frac{5\pi }{36}\right)}{2{\mathrm{cos}}^{2}\left(\frac{5\pi }{36}\right)-2i\mathrm{sin}\left(\frac{5\pi }{36}\right)\mathrm{cos}\left(\frac{5\pi }{36}\right)}\right)}^{3}$

=${\left(\frac{\mathrm{cos}\left(\frac{5\pi }{36}\right)+i\mathrm{sin}\left(\frac{5\pi }{36}\right)}{\mathrm{cos}\left(\frac{5\pi }{36}\right)-i\mathrm{sin}\left(\frac{5\pi }{36}\right)}\right)}^{3}$

=${\left(\frac{{e}^{i\frac{5\pi }{36}}}{{e}^{-i\frac{5\pi }{36}}}\right)}^{3}$

=${\left({e}^{i\frac{5\pi }{18}}\right)}^{3}$

=${e}^{i\frac{5\pi }{18}×3}$

=${e}^{i\frac{5\pi }{6}}$

=$\mathrm{cos}\frac{5\pi }{6}+i\mathrm{sin}\frac{5\pi }{6}$

=$-\frac{\sqrt{3}}{2}+\frac{i}{2}$
Let z be complex number such that $|\frac{z-i}{z+2i}|=1$ and |z| = $\frac{5}{2}$ . Then the value of |z + 3i| is :
1) $2\sqrt{3}$
2) $\sqrt{10}$
3) $\frac{15}{4}$
4) $\frac{7}{2}$
4
solution: ans: 4
$|\frac{z-i}{z+2i}|=1$

$⇒$|z – i| = |z + 2i|

(let z = x + iy) $⇒$x2 + (y – 1)2 = x2 + (y + 2)2

y = $-\frac{1}{2}$

Also given |z| = $\frac{5}{2}$

$⇒$x2 + y2 = $\frac{25}{4}$

$⇒$ x2 = 6

$\therefore$ z = $±\sqrt{6}$ $-\frac{1}{2}i$

|z + 3i| = $\sqrt{6+\frac{25}{4}}$ = $\frac{7}{2}$
If $\mathrm{Re}\left(\frac{z-1}{2z+i}\right)=1$ , where z = x + iy, then the point (x, y) lies on a:
1) straight line whose slope is $\frac{3}{2}$
2) straight line whose slope is $-\frac{2}{3}$
3) circle whose diameter is $\frac{\sqrt{5}}{2}$
4) circle whose centre is at $\left(-\frac{1}{2},-\frac{3}{2}\right)$
4
solution: ans: 3
$\mathrm{Re}\left(\frac{z-1}{2z+i}\right)=1$

Put z = x + iy $\therefore$ $\mathrm{Re}\left(\frac{\left(x+iy\right)-1}{2\left(x+iy\right)+i}\right)=1$

$⇒$

$\mathrm{Re}\left(\left(\frac{\left(x-1\right)+iy}{2x+i\left(2y+1\right)}\right)\left(\frac{2x-i\left(2y+1\right)}{2x-i\left(2y+1\right)}\right)\right)=1$

$⇒$

$\mathrm{Re}\left(\frac{\left\{\left(x-1\right)+iy\right\}\left\{2x-i\left(2y+1\right)\right\}}{4{x}^{2}+{\left(2y+1\right)}^{2}}\right)=1$

Real part of this equation is = 1 $\therefore$ $\frac{2x\left(x-1\right)+y\left(2y+1\right)}{4{x}^{2}+{\left(2y+1\right)}^{2}}$

$⇒$2x2 + 2y2 +2x + 3y + 1 = 0

This is an equation of circle. $\therefore$Locus is a circle whose center is $\left(-\frac{1}{2},-\frac{3}{4}\right)$ and radius $\frac{\sqrt{5}}{4}$

$\therefore$ Diameter = 2 $×$

$\frac{\sqrt{5}}{4}$

= $\frac{\sqrt{5}}{2}$
Let z $\in$ $\frac{2z-n}{2z+n}$ C with Im(z) = 10 and it satisfies = 2i - 1 for some natural number n. Then :
1) n = 20 and Re(z) = -10
2) n = 40 and Re(z) = 10
3) n = 40 and Re(z) = -10
4) n= 20 and Re(z) = 10
3
solution: ans: 3

Let Re (z) = x, then $\frac{2\left(x+10i\right)-n}{2\left(x+10i\right)+n}=2i-1$

$⇒\left(2x-n\right)+20i=-\left(2x+n\right)-40-20i+2ni$

$⇒2x-n=2x-n-40$

& 20 = -20 + 2n $⇒$x = -10 & n = 20
The equation |z – i| = |z – 1|, i = $\sqrt{-1}$ , represents :
1) a circle of radius of 1
2) the line through the origin with slope -1
3) a circle of radius $\frac{1}{2}$
4) the line through the origine with slope 1
4
solution: ans: 4

Let the complex number z = x + iy Now given | (x + iy) - 1 | = | (x+iy) - i | (x - 1)2 + y2 = x2 + (y - 1)2 x = y
If z and w are two complex numbers such that |zw| = 1 and arg(z) – arg(w) = $\frac{\pi }{2}$ , then:
1) $z\stackrel{―}{w}=\frac{1-i}{\sqrt{2}}$
2) $\stackrel{―}{z}w=i$
3) $z\stackrel{―}{w}=\frac{-1+i}{\sqrt{2}}$
4) $\stackrel{―}{z}w=-i$
4
solution: ans: 4 $|zw|=1$

let $w=\frac{1}{r}{e}^{i\theta }$

then z = $r{e}^{i\left(\theta +\frac{\pi }{2}\right)}$

$\stackrel{―}{z}w={e}^{-i\left(\theta +\frac{\pi }{2}\right)}.{e}^{i\theta }={e}^{-i\left(\pi /2\right)}=-i$

$z\stackrel{―}{w}={e}^{i\left(\theta +\frac{\pi }{2}\right)}.{e}^{-i\theta }={e}^{i\pi /2}=i$
If a > 0 and z = $\frac{{\left(1+i\right)}^{2}}{a-i}$ , has magnitude $\sqrt{\frac{2}{5}}$ , then $\stackrel{―}{z}$ is equal to :
1) $-\frac{1}{5}+\frac{3}{5}i$
2) $-\frac{1}{5}-\frac{3}{5}i$
3) $\frac{1}{5}-\frac{3}{5}i$
4) $-\frac{3}{5}-\frac{1}{5}i$
2
solution: ans: 2
$z=\frac{{\left(1+i\right)}^{2}}{a-i}×\frac{a+i}{a+i}$

$⇒z=\frac{\left(1-1+2i\right)\left(a+i\right)}{{a}^{2}+1}=\frac{2ai-2}{{a}^{2}+1}$

$⇒|z|=\sqrt{{\left(\frac{-2}{{a}^{2}+1}\right)}^{2}+{\left(\frac{2a}{{a}^{2}+1}\right)}^{2}}=\sqrt{\frac{4+4{a}^{2}}{{\left({a}^{2}+1\right)}^{2}}}$

$⇒\sqrt{\frac{4\left(1+{a}^{2}\right)}{{\left({a}^{2}+1\right)}^{2}}}=\frac{2}{\sqrt{{a}^{2}+1}}$ -------(1)

given $|z|=\sqrt{\frac{2}{5}}$

so $\sqrt{\frac{2}{5}}=\frac{2}{\sqrt{1+{a}^{2}}}$ from equation(1)

$⇒\frac{2}{5}=\frac{4}{1+{a}^{2}}$

$⇒1+{a}^{2}=10$

a = $±$ 3

$\because$ a is greater than 3 a = 3
$\stackrel{―}{z}=\frac{{\left(1-i\right)}^{2}}{3+i}$

$\frac{\left(1+{i}^{2}-2i\right)}{3+i}$

$\frac{\left(1+{i}^{2}-2i\right)\left(3-i\right)}{\left(3+i\right)\left(3-i\right)}$

$\frac{\left(-2i\right)\left(3-i\right)}{\left(9-{i}^{2}\right)}$

$\frac{\left(-2i\right)\left(3-i\right)}{10}$

$\frac{-6i+2{i}^{2}}{10}$

$\frac{-6i-2}{10}$

$-\frac{1}{5}-\frac{3}{5}i$

Let $\mathrm{z}=a+ib,b\ne 0$ be complex numbers satisfying ${z}^{2}=\overline{z}\cdot {2}^{1-z}$ . Then the least value of $n\in N$ , such that ${z}^{n}=\left(z+1{\right)}^{n}$ , is equal to __________.
6
solution: ans: 6

$\because$ ${z}^{2}=\stackrel{―}{z}\phantom{\rule{0.167em}{0ex}}.\phantom{\rule{0.167em}{0ex}}{2}^{1-|z|}$

$⇒|z{|}^{2}=|\stackrel{―}{z}|\phantom{\rule{0.167em}{0ex}}.\phantom{\rule{0.167em}{0ex}}{2}^{1-|z|}$ ----(1)

$⇒|z|={2}^{1-|z|}$

$\because$ $b\ne 0⇒|z|\ne 0$

$\therefore$ $|z|=1$ ---(2)

$\because$ $z=a+ib$ then $\sqrt{{a}^{2}+{b}^{2}}=1$ -----(3)

Now again from equation (1), equation (2), equation (3) we get : ${a}^{2}-{b}^{2}+i2ab=\left(a-ib\right){2}^{0}$

$\therefore$ ${a}^{2}-{b}^{2}=a$ and $2ab=-b$

$\therefore$ $a=-\frac{1}{2}$ and $b=\phantom{\rule{0.167em}{0ex}}±\phantom{\rule{0.167em}{0ex}}\frac{\sqrt{3}}{2}$

$z=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ or $z=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$

${z}^{n}=\left(z+1{\right)}^{n}⇒{\left(\frac{z+1}{z}\right)}^{n}=1$

${\left(1+\frac{1}{z}\right)}^{n}=1$

$\left(\frac{1+\sqrt{3}i}{2}\right)=1$

then the minimum value of n is 6
Let $S=\left\{z\in C:{z}^{2}+\overline{z}=0\right\}$ . Then $\sum _{z\in S}\left(\mathrm{Re}\left(z\right)+\mathrm{Im}\left(z\right)\right)$ is equal to ______________.
0
solution: ans = 0

$\because$ ${z}^{2}+\stackrel{―}{z}=0$

Let $z=x+iy$

$\therefore$ ${x}^{2}-{y}^{2}+2ixy+x-iy=0$

$\left({x}^{2}-{y}^{2}+x\right)+i\left(2xy-y\right)=0$

$\therefore$ ${x}^{2}+{y}^{2}=0$ and $\left(2x-1\right)y=0$

if $x=\phantom{\rule{0.167em}{0ex}}+\phantom{\rule{0.167em}{0ex}}\frac{1}{2}$ then $y=\phantom{\rule{0.167em}{0ex}}±\phantom{\rule{0.167em}{0ex}}\frac{\sqrt{3}}{2}$

and if $y=0$

then $x=0,-1$

$\therefore$ $z=0+0i,-1+0i,\frac{1}{2}+\frac{\sqrt{3}}{2}i,\frac{1}{2}-\frac{\sqrt{3}}{2}i$

$\therefore$ $\sum \left({R}_{e}\left(z\right)+m\left(z\right)\right)=0$
If ${z}^{2}+z+1=0$ , $z\in C$ , then$|\sum _{n=1}^{15}{\left({z}^{n}+{\left(-1\right)}^{n}\frac{1}{{z}^{n}}\right)}^{2}|$ is equal to _________.
2

$\because$ ${z}^{2}+z+1=0$

$⇒$ $\omega$ or $\omega$2

$\because$ $|\sum _{n=1}^{15}{\left({z}^{n}+{\left(-1\right)}^{n}\frac{1}{{z}^{n}}\right)}^{2}|$

$=|\sum _{n=1}^{15}{z}^{2n}+\sum _{n=1}^{15}{z}^{-2n}+2\phantom{\rule{0.167em}{0ex}}.\phantom{\rule{0.167em}{0ex}}\sum _{n=1}^{15}{\left(-1\right)}^{n}|$

$=|0+0-2|$

$=2$
Let $z=\frac{1-i\sqrt{3}}{2}$ , $i=\sqrt{-1}$ . Then the value of $21+{\left(z+\frac{1}{z}\right)}^{3}+{\left({z}^{2}+\frac{1}{{z}^{2}}\right)}^{3}+{\left({z}^{3}+\frac{1}{{z}^{3}}\right)}^{3}+....+{\left({z}^{21}+\frac{1}{{z}^{21}}\right)}^{3}$ is ______________.
13

$z=\frac{1-\sqrt{3i}}{2}={e}^{-i\frac{\pi }{3}}$

${z}^{r}+\frac{1}{{z}^{r}}=2\mathrm{cos}\left(-\frac{\pi }{3}\right)r=2\mathrm{cos}\frac{r\pi }{3}$

$⇒21+\sum _{r=1}^{21}{\left({z}^{r}+\frac{1}{{z}^{r}}\right)}^{3}=8\left({\mathrm{cos}}^{3}\frac{r\pi }{3}\right)=2\left(\mathrm{cos}r\pi +3\mathrm{cos}\frac{r\pi }{3}\right)$

$⇒21+{\left(z+\frac{1}{2}\right)}^{3}+{\left({z}^{2}+\frac{1}{{z}^{2}}\right)}^{3}+....{\left({z}^{21}+\frac{1}{{z}^{21}}\right)}^{3}$

$=21+\sum _{r=1}^{21}{\left({z}^{r}+\frac{1}{{z}^{r}}\right)}^{3}$

$=21+\sum _{r=1}^{21}\left(2\mathrm{cos}r\pi +6\mathrm{cos}\frac{r\pi }{3}\right)$

$=21-2-6$

$=13$
If the real part of the complex number $z=\frac{3+2i\mathrm{cos}\theta }{1-3i\mathrm{cos}\theta },\theta \in \left(0,\frac{\pi }{2}\right)$ is zero, then the value of sin 2 3 $\theta$ + cos 2 $\theta$ is equal to _______________.
1
solution: ans = 1
since Re(z) = 0, calculate the real part of the z and equate it to zero

$z=\frac{3+2i\mathrm{cos}\theta }{1-3i\mathrm{cos}\theta }.\frac{1+3i\mathrm{cos}\theta }{1+3i\mathrm{cos}\theta }$

on solving this we get $\theta$ = $\frac{\pi }{4}$ so sin 2 3 $\theta$ =1
Let z1, z2 be the roots of the equation z2 + az + 12 = 0 and z1, z2 form an equilateral triangle with origin. Then, the value of |a| is
6
solution: ans=6

For equilateral triangle with vertices 1, z2 and z2, ${z}_{1}^{2}+{z}_{2}^{2}+{z}_{3}^{3}={z}_{1}{z}_{2}+{z}_{2}{z}_{3}+{z}_{3}{z}_{1}$

Here one vertex z3 is 0

$\therefore$ ${z}_{1}^{2}+{z}_{2}^{2}={z}_{1}{z}_{2}+0+0$

${z}^{2}+az+12=0$

$\therefore$ ${z}_{1}+{z}_{2}=-a$

${z}_{1}{z}_{2}=12$

$\therefore$

${z}_{1}^{2}+{z}_{2}^{2}+2{z}_{1}{z}_{2}={z}_{1}{z}_{2}+2{z}_{1}{z}_{2}$

$⇒\left({z}_{1}+{z}_{2}{\right)}^{2}=3{z}_{1}{z}_{2}$

$⇒\left(-a{\right)}^{2}=3×12$

$⇒{a}^{2}=36$

$⇒a=±6$

$⇒|a|\phantom{\rule{0.167em}{0ex}}=6$
Let z and $\omega$ be two complex numbers such that $\omega =z\stackrel{―}{z}-2z+2,|\frac{z+i}{z-3i}|=1$ and Re($\omega$ ) has minimum value. Then, the minimum value of n $\in$ N for which $\omega$n n is real, is equal to ______________.
4
solution: ans = 4

Let z = x + iy | z + i | = | z $-$ 3i |

$⇒$ y = 1

$\omega$ = x2 + y2 - 2x - 2iy + 2

$\omega$ = x2 + 12 - 2x - 2i + 2

Re($\omega$) = x 2 - 2x + 3

Re($\omega$) = (x - 1) 2 + 2

Re($\omega$)min at x = 1 $⇒$ z = 1 + i

$⇒$ $\omega$ = 2(1 - i) = $\sqrt{2}{e}^{i\left(\frac{-\pi }{4}\right)}$

$\omega$n = $\sqrt{2}{e}^{i\left(\frac{-n\pi }{4}\right)}$

for $\omega$n to be real least value of n in N must be 4
If the least and the largest real values of a, for which the equation z + α |z – 1| + 2i = 0 (z ∈ C and i 2= - 1 ) has a solution, are p and q respectively; then 4(p2 + q2) is equal to ______.
10
solution: ans = 10

$x+iy+\alpha \sqrt{{\left(x-1\right)}^{2}+{y}^{2}}+2i=0$

$\therefore$ y + 2 =0 $x+\alpha \sqrt{{\left(x-1\right)}^{2}+{y}^{2}}=0$

y = $-$ 2 and

${x}^{2}={\alpha }^{2}\left({x}^{2}-2x+1+4\right)$

${\alpha }^{2}=\frac{{x}^{2}}{{x}^{2}-2x+5}⇒{x}^{2}\left({\alpha }^{2}-1\right)-2x{\alpha }^{2}+5{\alpha }^{2}=0$

$x\in R⇒D\ge 0$

$4{\alpha }^{4}-4\left({\alpha }^{2}-1\right)5{\alpha }^{2}\ge 0$

${\alpha }^{2}\left[4{\alpha }^{2}-2{\alpha }^{2}+20\right]\ge 0$

${\alpha }^{2}\left[-16{\alpha }^{2}+20\right]\ge 0$

${\alpha }^{2}\left[{\alpha }^{2}-\frac{5}{4}\right]\le 0$

$0\le {\alpha }^{2}\le \frac{5}{4}$

$\therefore$ ${\alpha }^{2}\in \left[0,\frac{5}{4}\right]$

$\therefore$ $\alpha \in \left[-\frac{\sqrt{5}}{2},\frac{\sqrt{5}}{2}\right]$

$4\left[\left(q{\right)}^{2}+\left(p{\right)}^{2}\right]=4\left[\frac{5}{4}+\frac{5}{4}\right]=10$

If ${\left(\frac{1+i}{1-i}\right)}^{\frac{m}{2}}={\left(\frac{1+i}{1-i}\right)}^{\frac{n}{3}}=1$ , (m, n $\in$ N) then the greatest common divisor of the least values of m and n is _______ .
4
${\left(\frac{1+i}{1-i}\right)}^{m/2}={\left(\frac{1+i}{1-i}\right)}^{n/3}=1$

$⇒{\left(\frac{{\left(1+i\right)}^{2}}{2}\right)}^{m/2}={\left(\frac{{\left(1+i\right)}^{2}}{-2}\right)}^{n/3}=1$

$⇒\left(i{\right)}^{m/2}=\left(-i{\right)}^{n/3}=1$

$⇒\frac{m}{2}=4{k}_{1}\phantom{\rule{0.167em}{0ex}}and\phantom{\rule{0.167em}{0ex}}\frac{n}{3}=4{k}_{2}$

$⇒$m = 8 k1 and n = 12 k2

Least value of m = 8 and n = 12.

$\therefore$

GCD (8, 12) = 4
Let $i=\sqrt{-1}$ . If $\frac{{\left(-1+i\sqrt{3}\right)}^{21}}{{\left(1-i\right)}^{24}}+\frac{{\left(1+i\sqrt{3}\right)}^{21}}{{\left(1+i\right)}^{24}}=k$ , and $n=\left[|k|\right]$ be the greatest integral part of | k |. Then $\sum _{j=0}^{n+5}{\left(j+5\right)}^{2}-\sum _{j=0}^{n+5}\left(j+5\right)$ is equal to _________.
310
Solution: ans = 310

$\left(1+i{\right)}^{2}=1+{i}^{2}+2i=1-1+2i=2i$

$\left(1-i{\right)}^{2}=1+{i}^{2}-2i=1-1-2i=-2i$

we know, $-\frac{1}{2}+\frac{i\sqrt{3}}{2}=\omega$

$⇒-1+i\sqrt{3}=2\omega$

and $-\frac{1}{2}-\frac{i\sqrt{3}}{2}={\omega }^{2}$

$⇒-1-i\sqrt{3}=2{\omega }^{2}$

$⇒1+i\sqrt{3}=-2{\omega }^{2}$

$K=\frac{{\left(-1+i\sqrt{3}\right)}^{21}}{{\left(1-i\right)}^{24}}+\frac{{\left(1+i\sqrt{3}\right)}^{21}}{{\left(1+i\right)}^{24}}$

$=\frac{{\left(2\omega \right)}^{21}}{{\left({\left(1-i\right)}^{2}\right)}^{12}}+\frac{{\left(-2\omega \right)}^{21}}{{\left({\left(1+i\right)}^{2}\right)}^{12}}$

$=\frac{{2}^{21}.{\omega }^{21}}{{\left(-2i\right)}^{12}}+\frac{{\left(-2\right)}^{21}{\left({\omega }^{2}\right)}^{21}}{{\left(2i\right)}^{12}}$

${\omega }^{3}=1$

${i}^{4}=1$

$=\frac{{2}^{21}}{{2}^{12}}-\frac{{2}^{21}}{{2}^{12}}=0$

$\therefore$ $n=\left[|K|\right]=\left[|0|\right]=0$

$\sum _{j=0}^{5}{\left(j+5\right)}^{2}-\sum _{j=0}^{5}\left(j+5\right)$

$\sum _{j=0}^{5}\left({j}^{2}+25+10j-j-5\right)$

$\sum _{j=0}^{5}\left({j}^{2}+9j+20\right)$

$\sum _{j=0}^{5}{j}^{2}+9\sum _{j=0}^{5}j+20\sum _{j=0}^{5}1$

$\frac{5×6×11}{6}+9\left(\frac{5×6}{2}\right)+20×6$

= 310