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

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.

• This is the point on the circle $|z-(3+3i)|=1$ that lies furthest from the origin. – saulspatz Nov 15 '18 at 16:58

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?
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 $$z=3+3i+e^{i\pi/4}=(3+\frac12\sqrt{2})+(3+\frac12\sqrt{2})i=\frac{6+\sqrt{2}}{2} + \frac{6+\sqrt{2}}{2}i$$.