If \frac{z-1}{z+1} is purely an imaginary number and is not

Joyce Smith

Joyce Smith

Answered question

2022-01-18

If z1z+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

Travis Hicks

Beginner2022-01-19Added 29 answers

Write z=x+iy. Then:
z1z+1=x1+iyx+1+iy
Multiply top and bottom by (x+1)iy:
(x1+iy)(x+1iy)(x+1)2+y2=(x(1iy))(x+(1iy))(x+1)2+y2=
x2(1iy)2(x+1)2+y2=x2(12iyy2)(x+1)2+y2=
x2+y21(x+1)2+y2i2y(x+1)2+y2
Since the real part is 0, x2+y2=1. The modulus of a complex number is just x2+y2, so the value of the modulus is 1.
otoplilp1

otoplilp1

Beginner2022-01-20Added 41 answers

Let z1z+1=ki where k is any nonzero real.
Solving for z ( using componendo-et-dividendo) we get z=ki+1ki1
Hence |z|=|ki+1||ki1|=1 (as both numerator and denominaor equals (1+k2)12) Hence |z|=1

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