# Complex Numbers Questions Answers [SLST 2016]

In this post, we will discuss the questions from the previous year’s SLST exam on complex numbers with answers. The problems with solutions on complex numbers are given below.

## Previous Years SLST Questions on Complex Analysis

Q1: If a+ib = $\dfrac{c+i}{c-i}$ where a, b,c are real numbers, then a2+b2=? [WB SLST 2016]

(A) (c2+1)/(c2-1) (B) -c2

(C) i (D) 1

a+ib = $\dfrac{c+i}{c-i}$ = $\dfrac{(c+i)^2}{(c-i)(c+i)}$ = $\dfrac{c^2-1+2ci}{c^2+1}$

Thus, a=$\dfrac{c^2-1}{c^2+1}$ and b=$\dfrac{2c}{c^2+1}$

⇒ a2+b2 = $(\dfrac{c^2-1}{c^2+1})^2$ + $(\dfrac{2c}{c^2+1})^2$

= $\dfrac{(c^2-1)^2+4c^2}{(c^2+1)^2}$

= $\dfrac{(c^2+1)^2}{(c^2+1)^2}$

= 1

So option(D) is correct.

Q2: For a complex number z, the number of solutions of the equation $z^2=\overline{z}$ is [WB SLST 2016]

(A) 1 (B) 3

(C) 4 (D) 2

The given equation is $z^2=\overline{z}$

Taking modulus on both sides, we get that

$|z|^2=|z|$ as $|\overline{z}|=|z|$

⇒ |z| = 0,1

Case 1: |z|=0 implies that z=0.

Case 2: |z|=1. So the given equation becomes

z2= $\overline{z}$ = z-1

⇒ z3=1

It has three solutions. So the total number of solutions of z2= z bar is equal to 4. So option(C) is correct.

Q3: If for a complex number z, |z|-z=1+2i, then the value of z is [WB SLST 2016]

(A) 2-$\frac{3}{2}$i (B) $\frac{3}{2}$-2i

(C) $\frac{3}{2}$+2i (D) $\frac{1}{2}$-2i

Let x=x+iy. Then |z| = $\sqrt{x^2+y^2}$.

|z|-z=1+2i

⇒ $\sqrt{x^2+y^2}$ – x-iy =1+2i

Comparing the real and imaginary parts of both sides, we get that

$\sqrt{x^2+y^2}$ – x=1 and y=-2

⇒ $\sqrt{x^2+4}$ – x=1 as y=-2.

⇒ x2+4 =1+2x+x2

⇒ 2x+1=4

⇒ x=3/2

∴ z =x+iy = 3/2 -2i.

So option(B) is correct.

Q4: The modulus (r) and the amplitude (θ) of the complex number z=1+i tan $\frac{3\pi}{5}$ are [WB SLST 2016]

(A) r=sec $\frac{3\pi}{5}$, θ=$\frac{3\pi}{5}$ (B) r=-sec $\frac{3\pi}{5}$, θ=-$\frac{2\pi}{5}$

(C) r=-sec $\frac{3\pi}{5}$, θ=$\frac{2\pi}{5}$ (D) r=-sec $\frac{3\pi}{5}$, θ=-$\frac{3\pi}{5}$

Let r(cosθ+isinθ) = 1+i tan $\frac{3\pi}{5}$.

Thus, rcosθ =1 and rsinθ=tan $\frac{3\pi}{5}$.

⇒ r2 = 1+tan2 $\frac{3\pi}{5}$ = sec2 $\frac{3\pi}{5}$

⇒ r = -sec $\frac{3\pi}{5}$ as sec $\frac{3\pi}{5}$<0.

So we deduce that

cosθ =-cos $\frac{3\pi}{5}$ and sinθ =-sin$\frac{3\pi}{5}$

⇒ θ = π+$\frac{3\pi}{5}$. As θ>π, this is not the principal argument.

⇒ arg z = θ-2π = -2π/5.

So option(B) is correct.

Q5: Find the principal value of ii. [WB SLST 2016]

We have i = cos(π/2)+isin(π/2) = eiπ/2.

Note that ii can be written as

ii = exp(i Log i)

= exp(i Log eiπ/2)

= exp( i(iπ/2) )

= exp(-π/2)

So the principal value of ii is equal to e-π/2.