If |z|, = 1,(z ne - 1)and z = x + iy,then lef | vedclass

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(b)$z = x + iy \Rightarrow |z{|^2} = {x^2} + {y^2} = 1$ .....$(i)$
Now, $\left( {\frac{{z - 1}}{{z + 1}}} \right) = \frac{{(x - 1) + iy}}{{(x + 1) +

Vedclass · Accepted Answer

Purely imaginary

Vedclass · Answer

(b)$z = x + iy \Rightarrow |z{|^2} = {x^2} + {y^2} = 1$ .....$(i)$
Now, $\left( {\frac{{z - 1}}{{z + 1}}} ight) = \frac{{(x - 1) + iy}}{{(x + 1) + iy}} 	imes \frac{{(x + 1) - iy}}{{(x + 1) - iy}}$
$ = \frac{{({x^2} + {y^2} - 1) + 2iy}}{{{{(x + 1)}^2} + {y^2}}}$$ = \frac{{2iy}}{{{{(x + 1)}^2} + {y^2}}}$ [by equation $(i)$]
Hence, $\left( {\frac{{z - 1}}{{z + 1}}} ight)$is purely imaginary.

If $|z|\, = 1,(z \ne - 1)$and $z = x + iy,$then $\left( {\frac{{z - 1}}{{z + 1}}} \right)$ is

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