If |z|, = 1,(z ne - 1) and z = x + iy, then left( {frac{{z

English

Gujarati

Hindi

4-1.Complex numbers

easy

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

Purely real

Purely imaginary

Zero

Undefined

Solution

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

Standard 11

Mathematics

Find the number of non-zero integral solutions of the equation $|1-i|^{x}=2^{x}$

medium

For any complex number $z,\bar z = \left( {\frac{1}{z}} \right)$if and only if

normal

If $z_1 = a + ib$ and $z_2 = c + id$ are complex numbers such that $| z_1 | = | z_2 |=1$ and $R({z_1}\overline {{z_2}} ) = 0$, then the pair of complex numbers $w_1 = a + ic$ and $w_2 = b + id$ satisfies

normal

If $z$ and $w$ are two complex numbers such that $|zw| = 1$ and $arg(z) -arg(w) =\frac {\pi }{2},$ then

hard

(JEE MAIN-2019)

Let $\alpha=8-14 i , A=\left\{ z \in C : \frac{\alpha z -\bar{\alpha} \overline{ z }}{ z ^2-(\overline{ z })^2-112 i }=1\right\}$ and $B =\{ z \in C 😐 z +3 i |=4\}$ Then $\sum_{z \in A \cap B}(\operatorname{Re} z-\operatorname{Im} z)$ is equal to $……………$.