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..
 A: 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 
A: $$\left|z+\frac{1}{z}\right|^2=\left|z+\frac{\bar z}{|z|^2}\right|^2 = \left|\frac 32\Re( z) -\frac i2\Im(z)\right|^2 = \frac{13}{4},$$
which leads to the system
$$\Re^2(z)+\Im^2(z)=2\\
\frac 94 \Re^2(z)+\frac14 \Im^2(z)=\frac{13}{4}.$$
This gives you $\Re^2(z)=\frac{11}{8}$ and $\Im^2(z)=\frac{5}{8}$. After that, given the argument of $z$, you say that $arg(z) = \pi - \arctan (\sqrt{5/11})$.
Therefore, $$z^{1991} = \left(\sqrt 2 \exp\left(i\pi-i\arctan \left (\sqrt{5/11 }\right)\right)\right)^{1991} = -2^{1991/2}\exp\left(-i\,1991\arctan \left (\sqrt{5/11 }\right)\right)$$
A: Assuming $z=-a+bi$ with $a,b\in\mathbb{R}^+$ so $a> 0$ and $b> 0$:


*

*The Absolute value:


$$\left|(-a+bi)+\frac{1}{(-a+bi)}\right|=\left|\frac{1+(-a+bi)^2}{(-a+bi)}\right|=\frac{|1+(-a+bi)^2|}{|(-a+bi)|}=\sqrt{\frac{(1+(-a)^2-b^2)^2+(2(-a)b)^2}{(-a)^2+b^2}}=\sqrt{\frac{(1+a^2-b^2)^2+(2ab)^2}{a^2+b^2}}$$


*The real and imaginary part:


$$\Re\left(-a+bi\right)^2+\Im\left(-a+bi\right)^2=(-a)^2+b^2=a^2+b^2$$


*The argument:


$$\arg\left(-a+bi\right)=\frac{\pi}{2}+\tan^{-1}\left(\frac{a}{b}\right)$$



*

*Solving $z$:
$$
\begin{cases}
\sqrt{\frac{(1+a^2-b^2)^2+(2ab)^2}{a^2+b^2}} = \frac{\sqrt{13}}{2} \\
a^2+b^2 = 2
\end{cases}\Longleftrightarrow
\begin{cases}
a=\frac{\sqrt{\frac{11}{2}}}{2},b=\pm\frac{\sqrt{\frac{5}{2}}}{2} \\
a=-\frac{\sqrt{\frac{11}{2}}}{2},b=\pm\frac{\sqrt{\frac{5}{2}}}{2}
\end{cases}
$$


But because $\arg(z)$ has to be between $\pi$ and $\frac{\pi}{2}$ the solutions that we are looking for are the ones with a positive $b$ (because the imaginary part has to be positive)! So we can use the following solution for $a$ and $b$:
$$a=-\frac{\sqrt{\frac{11}{2}}}{2} \space , \space b=\frac{\sqrt{\frac{5}{2}}}{2}$$

So:
$$z^{1991}=\left(-\frac{\sqrt{\frac{11}{2}}}{2}+\frac{\sqrt{\frac{5}{2}}}{2}i\right)^{1991}\approx -4.66055\cdot10^{299}-8.38854\cdot 10^{298}i$$
