<s>[INST] Find the complex number $z$ such that it satisfies: 1.$|z+\frac{1}{z}|=\frac{\sqrt{13}}{2} $, 2.$[Im (z)]^2+ [Re(z)]^2=2$ Find the complex number $z$ such that it satisfies: $$\left|z+\frac{1}{z}\right|=\frac{\sqrt{13}}{2} $$$$\Im (z)^2+ \Re(z)^2=2$$$$\frac{\pi}{2}<\arg(z)<\pi$$ then find $$z^{1991}$$
Know I was thinking that this second condition might mean whole part to the power of two, but am unsure about that, just wanted some input on how to do these types of assignments because we having done anything like this before..
 [/INST] Hint 
Let $z=a+bi$ then 
$$a^2+b^2=2,\dfrac{1}{z}=\dfrac{a-bi}{a^2+b^2}$$
and 
$$|z+\dfrac{1}{z}|^2=\left(a+\dfrac{a}{a^2+b^2}\right)^2
+\left(b-\dfrac{b}{a^2+b^2}\right)^2=\dfrac{13}{4}$$
so
$$(a+\frac{a}{2})^2+(b-\frac{b}{2})^2=\frac{13}{4}$$
so
$$9a^2+b^2=13$$
then
$$a^2=\dfrac{11}{8},b^2=\dfrac{5}{8}$$
so
$$z=\sqrt{a^2+b^2}e^{ix}=\sqrt{2}e^{ix}$$ then you can do it 
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