If z0 is a complex number such that |z-1z|=2, then the maximum value of |z| is:

Question

If $z
eq0$ is a complex number such that $\left|z-\dfrac1zight|=2$, then the maximum value of $|z|$ is:

Accepted Answer

Let $z=re^{i\theta}$. Then:
$$z-\frac1z = \left(r-\frac1r\right)\cos\theta+i\left(r+\frac1r\right)\sin\theta.$$

$$\left|z-\frac1z\right|^2=\left(r-\frac1r\right)^2+\left[\left(r+\frac1r\right)^2-\left(r-\frac1r\right)^2\right]\sin^2\theta = \left(r-\frac1r\right)^2+4\sin^2\theta = 4.$$

For a solution $\theta$ to exist, we need $\left(r-\dfrac1r\right)^2\le4$, i.e. $\left|r-\dfrac1r\right|\le2$. The maximum $r$ satisfying this occurs at equality (for $r>1$):
$$r-\frac1r=2 \implies r^2-2r-1=0 \implies r=1+\sqrt2 \ (\text{taking the positive root}).$$

Solutions · 1