This "If z−1z+1 is purely an imaginary number and..." has been explained

Joyce Smith Answered2022-01-18

\frac{z - 1}{z + 1}

is purely an imaginary number and is not equal to -1, then what is the value of the modulus of z?

Answer & Explanation

Travis Hicks

Expert

2022-01-19Added 29 answers

Write

z = x + i y

. Then:

\frac{z - 1}{z + 1} = \frac{x - 1 + i y}{x + 1 + i y}

Multiply top and bottom by

(x + 1) - i y

\frac{(x - 1 + i y) (x + 1 - i y)}{{(x + 1)}^{2} + y^{2}} = \frac{(x - (1 - i y)) (x + (1 - i y))}{{(x + 1)}^{2} + y^{2}} =

\frac{x^{2} - {(1 - i y)}^{2}}{{(x + 1)}^{2} + y^{2}} = \frac{x^{2} - (1 - 2 i y - y^{2})}{{(x + 1)}^{2} + y^{2}} =

\frac{x^{2} + y^{2} - 1}{{(x + 1)}^{2} + y^{2}} - i \frac{2 y}{{(x + 1)}^{2} + y^{2}}

Since the real part is 0,

x^{2} + y^{2} = 1

. The modulus of a complex number is just

\sqrt{x^{2} + y^{2}}

, so the value of the modulus is 1.

otoplilp1

Expert

2022-01-20Added 41 answers

Let

\frac{z - 1}{z + 1} = k i

where k is any nonzero real.
Solving for z ( using componendo-et-dividendo) we get

- z = \frac{k i + 1}{k i - 1}

Hence

| z | = \frac{| k i + 1 |}{| k i - 1 |} = 1

(as both numerator and denominaor equals

{(1 + k^{2})}^{\frac{1}{2}}

) Hence

| z | = 1

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