Joyce Smith

2022-01-18

If $\frac{z-1}{z+1}$ is purely an imaginary number and is not equal to -1, then what is the value of the modulus of z?

Travis Hicks

Expert

Write $z=x+iy$. Then:
$\frac{z-1}{z+1}=\frac{x-1+iy}{x+1+iy}$
Multiply top and bottom by $\left(x+1\right)-iy$:
$\frac{\left(x-1+iy\right)\left(x+1-iy\right)}{{\left(x+1\right)}^{2}+{y}^{2}}=\frac{\left(x-\left(1-iy\right)\right)\left(x+\left(1-iy\right)\right)}{{\left(x+1\right)}^{2}+{y}^{2}}=$
$\frac{{x}^{2}-{\left(1-iy\right)}^{2}}{{\left(x+1\right)}^{2}+{y}^{2}}=\frac{{x}^{2}-\left(1-2iy-{y}^{2}\right)}{{\left(x+1\right)}^{2}+{y}^{2}}=$
$\frac{{x}^{2}+{y}^{2}-1}{{\left(x+1\right)}^{2}+{y}^{2}}-i\frac{2y}{{\left(x+1\right)}^{2}+{y}^{2}}$
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

Let $\frac{z-1}{z+1}=ki$ where k is any nonzero real.
Solving for z ( using componendo-et-dividendo) we get $-z=\frac{ki+1}{ki-1}$
Hence $|z|=\frac{|ki+1|}{|ki-1|}=1$ (as both numerator and denominaor equals ${\left(1+{k}^{2}\right)}^{\frac{1}{2}}$) Hence $|z|=1$

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