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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.

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2 Answers 2

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

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  • $\begingroup$ okay I see that now, how did you see that? I wasn't even looking for that. $\endgroup$
    – yoshi
    Commented Sep 29, 2018 at 2:59
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    $\begingroup$ @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. $\endgroup$
    – dxiv
    Commented Sep 29, 2018 at 3:01
  • $\begingroup$ Also note $z/2$ and $2/z$ are complex cube root of unity. So $z/2 +2/z=-1$ $\endgroup$
    – user585765
    Commented Sep 30, 2018 at 8:10
  • $\begingroup$ @LoopBack Right, $\,z/2+2/z=z/2+\overline{z/2}=2 \operatorname{Re}(z/2)\,$. $\endgroup$
    – dxiv
    Commented Sep 30, 2018 at 16:06
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$$z = -1 + \sqrt{3}i=2e^{i2\pi /3} $$ It depends on how experienced you are in working with polar forms of complex numbers.

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  • $\begingroup$ Aye - ya I bungled the conversion too $\endgroup$
    – yoshi
    Commented Sep 29, 2018 at 3:09

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