Let $z$ and $\omega$ be two complex numbers such that $\omega=z \bar{z}-2 z+2,\left|\frac{z+i}{z-3 i}\right|=1$ and $\operatorname{Re}(\omega)$ has minimum value. Then, the minimum value of $\mathrm{n} \in \mathrm{N}$ for which $\omega^{\mathrm{n}}$ is real, is equal to

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Complex Numbers

Let

z

and

ω

be two complex numbers such that

ω = z \overset{z}{ˉ} - 2 z + 2, \frac{z + i}{z - 3 i} = 1

and

Re (ω)

has minimum value. Then, the minimum value of

n \in N

for which

ω^{n}

is real, is equal to

Answer: 4

The condition

\frac{z + i}{z - 3 i} = 1

means

∣ z + i ∣ = ∣ z - 3 i ∣

, so

z

lies on the perpendicular bisector of the points

- i

and

3 i

, which is the line

Im z = 1

. Putting

z = x + i

ω = z \overset{z}{ˉ} - 2 z + 2 = (x^{2} + 1) - 2 (x + i) + 2 = x^{2} - 2 x + 3 - 2 i

Re ω = (x - 1)^{2} + 2

, which is least at

x = 1

, where

ω = 2 - 2 i

. In polar form

ω = 22 (cos \frac{π}{4} - i sin \frac{π}{4})

ω^{n}

has argument

- \frac{nπ}{4}

, and this is a multiple of

π

for the first time at

n = 4

. The least such natural number is

4

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Integer answer