Question

A function $f$ is defined on the complex number by $f(z)=(a+bi)z$, where ' $a$ ' and 'b' are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin. Given that $|a+bi|=8$ and that $b^2=\frac{u}{v}$ where $u$ and $v$ are coprimes. Find the value of $(u+v)$.

Answer 1

Answer: 259

Given $|(a+bi)z-z|=|(a+bi)z|$
$|z(a-1)+biz|=|az+bzi|$
$|z||(a-1)+bi|=|z||a+bi|$
$\therefore (a-1{)}^2+b^2=a^2+b^2$
$\therefore a=1/2$
$\text{ since }|a+bi|=8$
$a^2+b^2=64$
$b^2=64-\frac{1}{4}=\frac{255}{4}=\frac{u}{v}$
$\therefore u=255\text{ and }v=4$
$\therefore u+v=259$