Determine the complex number Z that satisfy $|z-3-3i|=1$ and that has maximum absolute value

Question

I'm having a hard time solving following question:

Determine the complex number Z that satisfy $|z-3-3i|=1$ and that has maximum absolute value.

Z should be written on the form z=x+iy.

I have determined with the help of the triangle inequality that $|z|=1+\sqrt{18}$.

This is the point where i run into problem. I don't know how to determine z on the form $z=x+iy$, using the information above.

If someone could give me a hint i would be very thankful.

Answer 1

Think of the complex numbers as a real coordinate plane. The equation $|z-3-3i|=1$ is basically a circle of radius $1$ with the center at $(3, 3)$. What point on the circumference is farthest away from the origin?

Answer 2

Any number with magnitude $1$ can be written as $e^{it}$ for real $t$, so your $z$ must be of the form $z= z_0+e^{it}$ where $z_0=3+3i$.

Note that we can rewrite $z_0$ as $3\sqrt{2}e^{i\pi/4}$. Let $\mathbf v_0$ denote the vector from $0$ to $z_0$, and $\mathbf v_1$ the vector from $0$ to $e^{it}$. You are looking to maximize $|\mathbf v_0+\mathbf v_1|$.

The vector $\mathbf v_0$ always points in the direction of the angle $\pi/4$. As we let $t$ vary, $\mathbf v_1$ swings around, pointing in the direction of the angle $t$. The sum of $\mathbf v_0$ and $\mathbf v_1$ has the greatest magnitude (namely $|\mathbf v_0| + |\mathbf v_1|$) when the two vectors point in the same direction. That is, when $t=\pi/4$.

So $|z|$ is maximized by taking $z=3+3i+e^{i\pi/4}=(3+\frac12\sqrt{2})+(3+\frac12\sqrt{2})i=\frac{6+\sqrt{2}}{2} + \frac{6+\sqrt{2}}{2}i$.