Let z satisfy |z|=1 and z=1-z.

Question

Let $z$ satisfy $|z|=1$ and $\bar z=1-z$.

Statement 1: $z$ is a real number.

Statement 2: Principal argument of $z$ is $\dfrac\pi3$.

Accepted Answer

Let $z=x+iy$. $\bar z=1-z \implies x-iy=1-x-iy \implies 2x=1 \implies x=\dfrac12$. From $|z|=1$: $y^2=1-\dfrac14=\dfrac34 \implies y=\pm\dfrac{\sqrt3}2$.

$	an	heta=\dfrac{\sqrt3/2}{1/2}=\sqrt3 \implies 	heta=\dfrac\pi3$ (principal argument). Since $y
e0$, $z$ is not real.

So statement 1 is false, statement 2 is true.

Solutions · 1