If `z=(1+7i)/((2-i)^2)` , then (a)`|z|=2` (b) `|z|=1/2` c) `a m p(z)=pi/4` (d) `a m p(z)=(3pi)/4`

Explanation

We have given,

`z=frac{(1+7i)}{(2-i)^2}`

`z=frac{(1+7i)}{(4+i^2-4i)}`

`z=frac{(1+7i)}{(3-4i)}`

`z=frac{(1+7i)(3+4i)}{(3-4i)(3+4i)}`

`z=frac{25(-1+i)}{25}`

`z=-1+i`

Now `\tan \alpha =\ | \frac{Im(z)}{Re(z)} \ |`

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