If a>0 and z=\frac{(1+i)^{2}}{a-i} , has magnitude \sqrt{\frac{2}{5}} ,

then \bar{z} is equal to :

  • Option 1)

    -\frac{1}{5}-\frac{3}{5}i

  • Option 2)

    -\frac{3}{5}-\frac{1}{5}i

  • Option 3)

    \frac{1}{5}-\frac{3}{5}i

  • Option 4)

    -\frac{1}{5}+\frac{3}{5}i

 

Answers (1)

z=\frac{(1+i)^{2}}{a-i}

    =\frac{1+2i-1}{a-i}=\frac{2i}{a-i}

z=\frac{2i}{a-i}\times \frac{a+i}{a+i}=\frac{2i(a+i)}{a^{2}+1}

given that 

|z|=\sqrt{\frac{2}{5}}

|\frac{2i(a+i)}{a^{2}+1}|=\sqrt{\frac{2}{5}}

\frac{2}{\sqrt{a^{2}+1}}=\sqrt{\frac{2}{5}}

a=3

z=-\frac{1}{5}-\frac{3i}{5}

Option (1) is correct.


Option 1)

-\frac{1}{5}-\frac{3}{5}i

Option 2)

-\frac{3}{5}-\frac{1}{5}i

Option 3)

\frac{1}{5}-\frac{3}{5}i

Option 4)

-\frac{1}{5}+\frac{3}{5}i

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