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 a ^{2}+b^{2}=? [WB SLST 2016] **

**(A)** (c^{2}+1)/(c^{2}-1) **(B)** -c^{2}

**(C)** i **(D)** 1

**Answer:**

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

⇒ a^{2}+b^{2} = $(\dfrac{c^2-1}{c^2+1})^2$ + $(\dfrac{2c}{c^2+1})^2$

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

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

= 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

**Answer:**

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

z^{2}= $\overline{z}$ = z^{-1}

⇒ z^{3}=1

It has three solutions. So the total number of solutions of z^{2}= 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

**Answer:**

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.

⇒ x^{2}+4 =1+2x+x^{2}

⇒ 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=-$\frac{3\pi}{5}$

**Answer:**

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

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

⇒ r^{2} = 1+tan^{2} $\frac{3\pi}{5}$ = sec^{2} $\frac{3\pi}{5}$

⇒ 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 i^{i}.**

**[WB SLST 2016]**

**Answer:**

We have i = cos(π/2)+isin(π/2) = e^{iπ/2}.

Note that i^{i} can be written as

i^{i} = exp(i Log i)

= exp(i Log e^{iπ/2})

= exp( i(iπ/2) )

= exp(-π/2)

So the principal value of i^{i} is equal to e^{-π/2}.

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