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

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

The values of $z$for which $|z + i|\, = \,|z - i|$ are

Let $z_1, z_2 \in C$ such that $| z_1 + z_2 |= \sqrt 3$ and $|z_1| = |z_2| = 1,$ then the value of $|z_1 - z_2|$ is

The modulus and amplitude of $\frac{{1 + 2i}}{{1 - {{(1 - i)}^2}}}$ are

Let $z$be a purely imaginary number such that ${\mathop{\rm Im}\nolimits} \,(z) > 0$. Then $arg(z)$ is equal to

Let $z$ =${i^{2i}}$ , then $|z|$ is (where $i$ =$\sqrt { - 1}$ )