If z -1- z -1 is purely imaginary number z -1- find the value of - z -

Let z = x + iy, z 1/z+1 =x+iy 1/x+iy+1=x 1+iy/x+1+iy =(x 1+iy)(x+1 iy)/(x+1+iy)(x+1 iy) [Rationalizing the denominator] =(x 1+iy)(x+1 iy)/(x+1)^2 (iy)^2 ...

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Byju's Answer

Standard XII

Mathematics

Purely Imaginary

If z -1/ z +1...

Question

If $\frac{z - 1}{z + 1}$ is purely imaginary number $(z \neq - 1)$ , find the value of |z|.

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Solution

Let z = x + iy,

$\frac{z - 1}{z + 1} = \frac{x + i y - 1}{x + i y + 1} = \frac{x - 1 + i y}{x + 1 + i y} = \frac{(x - 1 + i y) (x + 1 - i y)}{(x + 1 + i y) (x + 1 - i y)} [Rationalizing the denominator] = \frac{(x - 1 + i y) (x + 1 - i y)}{(x + 1)^{2} - (i y)^{2}} = \frac{x^{2} + x - i x y - x - 1 + i y + i x y + i y + y^{2}}{x^{2} + 2 x + 1 + y^{2}} = \frac{x^{2} - 1 + 21 y + y^{2}}{x^{2} + 2 x + 1 + y^{2}} = \frac{x^{2} + y^{2} - 1}{x^{2} + 2 x + 1 + y^{2}} + \frac{i 2 y}{x^{2} + 2 x + 1 + y^{2}}$

$∵$ It is purely imaginary number

$∴$ real part = 0

$= \frac{x^{2} + y^{2} - 1}{x^{2} + 2 x + 1 + y^{2}} = 0 = x^{2} + y^{2} - 1 = 0 = x^{2} + y^{2} = 1 = \sqrt{x^{2} + y^{2}} = 1 | z | = 1$

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If z−1z+1 is purely imaginary number (z≠−1), find the value of |z|.

If $\frac{z - 1}{z + 1}$ is purely imaginary number $(z \neq - 1)$ , find the value of |z|.