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If a>0 and z= (1+i)^2 a-i , has magnitude √ 2 5 , then overset- z is equal to:

JEE Main 2019 — Mathematics Algebra

2019mcqmedium

If a>0a>0a>0 and z=(1+i)2a−iz=\frac{{(1+i)}^{2}}{a-i}z=a−i(1+i)2​ , has magnitude 25\sqrt{\frac{2}{5}}52​​ , then z−\overset{-}{z}z− is equal to:

Official previous-year question

Held on 10 Apr 2019 · Verified 6 Jul 2026.

Options

  1. A

    −15−35i-\frac{1}{5}-\frac{3}{5}i−51​−53​i

  2. B

    −35−15i-\frac{3}{5}-\frac{1}{5}i−53​−51​i

  3. C

    15−35i\frac{1}{5}-\frac{3}{5}i51​−53​i

  4. D

    −15+35i-\frac{1}{5}+\frac{3}{5}i−51​+53​i

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