Let z be the complex number satisfying |z-5| 3 and having maximum positive principal argum…

Question

Let $z$ be the complex number satisfying $|z-5|\le 3$ and having maximum positive principal argument.
Then $34{|\frac{5z-12}{5iz+16}|}^{2}$ is equal to:

Accepted Answer

The region $|z-5|\le3$ is a disk centered at $(5,0)$ with radius $3$. The point of maximum argument lies where the line from the origin is tangent to the boundary circle. Since the distance from the origin to the center is $5$ and the radius is $3$,
$$\sin\theta=\frac35,\quad \cos\theta=\frac45,\quad |z|=\sqrt{5^2-3^2}=4.$$

So $z = 4(\cos\theta+i\sin\theta)=4\left(\dfrac45+i\dfrac35\right)=\dfrac{16}{5}+\dfrac{12i}{5}$, i.e. $5z=16+12i$.

Then:
$$\frac{5z-12}{5iz+16}=\frac{4+12i}{4+16i}=\frac{1+3i}{1+4i}.$$

$$\left|\frac{1+3i}{1+4i}\right|^2=\frac{1+9}{1+16}=\frac{10}{17}.$$

$$34\cdot\frac{10}{17}=20.$$

Solutions · 1