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(A) Let $z = r(\cos 	heta + i \sin 	heta)$,so $\bar{z} = r(\cos 	heta - i \sin 	heta)$.Then $(\bar{z})^2 + \frac{1}{z^2} = r^2(\cos 2	heta - i \s

Vedclass · Accepted Answer

$\left(\frac{43+3 \sqrt{205}}{2}\right)^{\frac{1}{4}}$

Vedclass · Answer

(A) Let $z = r(\cos 	heta + i \sin 	heta)$,so $\bar{z} = r(\cos 	heta - i \sin 	heta)$.Then $(\bar{z})^2 + \frac{1}{z^2} = r^2(\cos 2	heta - i \sin 2	heta) + \frac{1}{r^2}(\cos 2	heta - i \sin 2	heta) = (r^2 + \frac{1}{r^2}) \cos 2	heta - i (r^2 + \frac{1}{r^2}) \sin 2	heta$.Let $(r^2 + \frac{1}{r^2}) \cos 2	heta = m$ and $(r^2 + \frac{1}{r^2}) \sin 2	heta = -n$,where $m, n \in \mathbb{Z}$.Squaring and adding,we get $(r^2 + \frac{1}{r^2})^2 = m^2 + n^2$.Expanding,$r^4 + \frac{1}{r^4} + 2 = m^2 + n^2$.For option $(A)$,$|z|^4 = \frac{43+3 \sqrt{205}}{2}$.Then $r^4 + \frac{1}{r^4} + 2 = \frac{43+3 \sqrt{205}}{2} + \frac{2}{43+3 \sqrt{205}} + 2 = \frac{43+3 \sqrt{205}}{2} + \frac{43-3 \sqrt{205}}{22} + 2$ (Correction: $r^4 + \frac{1}{r^4} = \frac{43+3 \sqrt{205}}{2} + \frac{43-3 \sqrt{205}}{2} = 43$).Thus,$r^4 + \frac{1}{r^4} + 2 = 43 + 2 = 45$,which is $m^2 + n^2 = 45$. This is possible for integers $m, n$ (e.g.,$6^2 + 3^2 = 45$).Thus,option $(A)$ is correct.

List-$I$	List-$II$
$(P)$ $\|z\|^2$ is equal to	$(1)$ $12$
$(Q)$ $\|z-\bar{z}\|^2$ is equal to	$(2)$ $4$
$(R)$ $\|z\|^2+\|z+\bar{z}\|^2$ is equal to	$(3)$ $8$
$(S)$ $\|z+1\|^2$ is equal to	$(4)$ $10$
	$(5)$ $7$

Let $\bar{z}$ 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 $(\bar{z})^2+\frac{1}{z^2}$ are integers,then which of the following is/are possible value$(s)$ of $|z|$?

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