a function defined on the complex numbers

Question

A function $f$ is defined on the complex numbers by $$f(z) = (a + bi)z,$$where $a$ and $b$ are positive real 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 $$|a + bi| = 3,$$find $a$ and $b.$

So I've been trying to solve this and came up with this: The distance between $(a+bi)z$ and $z$ is $|(a+bi)z-z| = |(a+bi-1)z| = |(a+bi-1)|*|z|$. The distance between $(a+bi)z$ and the origin is simply $|(a+bi)z|$, but since $|a+bi| = 3$, $|(a+bi)z| = 3|z| = 3\sqrt{z\overline{z}}$. The function is said to have the property that thre image of each point in the complex plane is equidistant from that point and the origin. Therefore, $3|z| = |(a+bi-1)|*|z|$. Cancelling $|z|$ from both sides, $|a+bi-1| = 3$. Therefore, $a+bi-1 = 3$ or $a+bi-1 = -3$. The first case gives that $(a,b) = (4,0)$ while the latter case gives that $(a,b) = (-2,0)$.

But I realised that a and b must both be positive, rendering both my 'solution points' invalid. Where did I go wrong?

Answer 1

$f(z) = (a + bi)z$

$$|f(z)-z|=|f(z)|\land |a + bi| = 3$$

Let $z=x+iy,\;x,y\in\mathbb{R}$

We have $$|(a+bi)(x+iy)-(x+iy)|=|(a+bi)(x+iy)|\land |a+bi|=3$$ that is $$|a x-b y-x +i (a y+b x-y)| =|a x-b y +i (a y+b x)|\land |a+bi|=3$$

$$(a x-b y-x)^2+(a y+b x-y)^2=x^2+y^2\land a^2+b^2=9$$

$$\sqrt{(a y+b x-y)^2+(a x-b y-x)^2}=\sqrt{a^2+b^2} \sqrt{x^2+y^2}\land a^2+b^2=9$$

$$\left(a^2-2 a+b^2\right) \left(x^2+y^2\right)=0\land a^2+b^2=9$$

$$(a y+b x-y)^2+(a x-b y-x)^2=\left(a^2+b^2\right) \left(x^2+y^2\right)\land a^2+b^2=9$$ $$-2 a x^2-2 a y^2+x^2+y^2=0\land a^2+b^2=9$$

$$(1-2 a) \left(x^2+y^2\right)=0\land a^2+b^2=9$$

$$1-2a=0\to a=\frac12\to b^2=9-\frac14\to b=\pm \frac{\sqrt{35}}{2}$$

Solutions are $$\color{red}{a=\frac12;\;b=\pm \frac{\sqrt{35}}{2}}$$

Hope it is useful