Alternate Way of Computing Complex Polynomial?

I'm computing the value of this polynomial:

$$\left(\frac{2}{z}+\frac{z}{2}\right)^2+2$$

Where $$z = -1 + \sqrt{3}i$$

I converted to polar $$z = 4e^{i5\pi/6}$$ to grind out the first term then converted back to cartesian so I could add. I was wondering if there was a faster way of simplifying this expression. I ask because its a question on an old timed exam.

Hint:   note that $$\,\dfrac{z}{2}\,$$ is a complex cube root of unity, and $$\,\dfrac{2}{z}\,$$ its conjugate.

• okay I see that now, how did you see that? I wasn't even looking for that. Commented Sep 29, 2018 at 2:59
• @yoshi Once you've seen $\,e^{i \pi /4}=(1+i)/\sqrt{2}\,$, or $\,e^{i 2 \pi /3} = (-1 + i \sqrt{3})/2\,$ a few times, you'll tend to recognize them.
– dxiv
Commented Sep 29, 2018 at 3:01
• Also note $z/2$ and $2/z$ are complex cube root of unity. So $z/2 +2/z=-1$
– user585765
Commented Sep 30, 2018 at 8:10
• @LoopBack Right, $\,z/2+2/z=z/2+\overline{z/2}=2 \operatorname{Re}(z/2)\,$.
– dxiv
Commented Sep 30, 2018 at 16:06

$$z = -1 + \sqrt{3}i=2e^{i2\pi /3}$$ It depends on how experienced you are in working with polar forms of complex numbers.

• Aye - ya I bungled the conversion too Commented Sep 29, 2018 at 3:09