Question

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Answer 1

Answer:

Step-by-step explanation:

09

Answer 2

★Question:-

➠If z-1/z+1 is a purely imaginary number (z≠−1), then find the value of |z|.

★Answer:-

Let z = x + iy.Then,

→ z - 1 = x+iy - 1

→ z + 1 = x+iy + 1

$\\ \displaystyle \sf :\implies \frac{z-1}{z+1} =\frac{x+iy-1}{x+iy+1} \\\\\\:\implies \sf \frac{(x-1)+iy}{(x+1)+iy}$

Rationalising,

$\displaystyle :\implies \sf \frac{(x-1)+iy}{(x+1)+iy} \times \frac{(x+1)-iy}{(x+1)-iy} \\\\\\ \: (i^2 = -1)\\\\:\implies \sf \frac{(x-1)(x+1)-iy(x-1)+iy(x+1)+y^2}{(x+1)^2 -(iy)^2} \\\\\\:\implies \sf \frac{x^2- 1 \cancel{-xiy} -iy\cancel{+xiy}+iy+y^2}{(x+1)^2 +y^2} \\\\\\:\implies \sf \frac{(x^2 +y^2 -1)+i(2y)}{(x+1)^2+y^2} +\frac{2iy}{(x+1)^2 +y^2} \longrightarrow (1)$

For a purely imaginary number

→ z = a + bi

→z = bi

Where, a = 0

Comparing this with equation (1),

$\displaystyle :\implies \sf \frac{x^2 +y^2 - 1}{(x+1)^2 +y^2} = 0 \\\\\\:\implies \sf x^2 +y^2 - 1 = 0\\\\\\:\implies \sf x^2 +y^2 = 1$

Putting square roots on both sides,

$:\implies \sf \sqrt{x^2 +y^2} = \sqrt{1}\\\\\\:\implies \sf \sqrt{x^2 + y^2} = 1$

We know,

→ |z|=|x + iy|

     = √x² + y²

Therefore,

$:\implies \sf \sqrt{x^2 +y^2} = 1\\\\:\implies \underline{\boxed{\sf |z| = 1}}$

Hence,

The value of |z| is 1.

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