JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Complex Numbers

Let

\overset{z}{ˉ}

denote the complex conjugate of a complex number

z

. If

z

is a non-zero complex number for which both real and imaginary parts of

(\overset{z}{ˉ})^{2} + \frac{1}{z ^{2}}

are integers, then which of the following is/are possible value(s) of

∣ z ∣

Answer: (A)

Solution

Let

(\overset{z}{ˉ})^{2} + \frac{1}{z ^{2}} = m + in

where

m, n \in Z

.

Writing

z = x + i y

, we get:

(\overset{z}{ˉ})^{2} + \frac{z ˉ ^{2}}{∣ z ∣ ^{4}} = m + in

This leads to:

(1 + \frac{1}{∣ z ∣ ^{4}})^{2} ∣ z ∣^{4} = m^{2} + n^{2}

∣ z ∣^{4} + \frac{1}{∣ z ∣ ^{4}} + 2 = m^{2} + n^{2}

For option (A):

∣ z ∣^{4} = \frac{43 + 3 205}{2}

∣ z ∣^{4} + \frac{1}{∣ z ∣ ^{4}} = \frac{43 + 3 205}{2} + \frac{43 - 3 205}{2} = 43

m^{2} + n^{2} = 45

This gives

m = \pm 6, n = \pm 3

, which are integers. ✓

For options (B), (C), (D), similar calculations show that

m^{2} + n^{2}

is not an integer or doesn't yield integer values for

m

and

n

.

Therefore, only (A) is correct.

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · One or more correct options