JEE Advanced 2022 — Paper 2 — MATH-II solution

JEE Advanced2022Paper 2MATH-II

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 $∣ z ∣$ ?

A
$(\frac{43 + 3 205}{2})^{1/4}$
B
$(\frac{7 + 33}{4})^{1/4}$
C
$(\frac{9 + 65}{4})^{1/4}$
D
$(\frac{7 + 13}{6})^{1/4}$

Solution

Write $z = r e^{i θ}$ , so $\overset{z}{ˉ}^{2} = r^{2} e^{- 2 i θ}$ and $\frac{1}{z ^{2}} = \frac{1}{r ^{2}} e^{- 2 i θ}$ . Therefore

$(\overset{z}{ˉ})^{2} + \frac{1}{z ^{2}} = (r^{2} + \frac{1}{r ^{2}}) e^{- 2 i θ} = a + ib$ , with $a, b \in Z$ .

Taking moduli: $(r^{2} + \frac{1}{r ^{2}})^{2} = a^{2} + b^{2}$ .

Set $u = r^{4}$ . Then $u^{2} - (a^{2} + b^{2} - 2) u + 1 = 0$ , giving $u = \frac{( a ^{2} + b ^{2} - 2 ) \pm ( a ^{2} + b ^{2} - 2 ) ^{2} - 4}{2}$ .

For integer $(a, b)$ with $a^{2} + b^{2} = 45$ (such as $(6, 3)$ , $(3, 6)$ and sign variants), one obtains $u = \frac{43 + 3 205}{2}$ , hence $∣ z ∣ = u^{1/4} = (\frac{43 + 3 205}{2})^{1/4}$ .

The other options do not arise from integer values of $a^{2} + b^{2}$ in this manner, so only (A) is valid.

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