<s>[INST] Express the value $z$ below in polar form, and the value $w$ in the form $a+bi$. I have been having a lot of issues on determining how to work through problems of the sort and I would be very grateful if somebody could provide me with a guided/ explained answer to enable me to understand how to work through these.
Given:
Picture of Question
Express the value $z$ below in polar form, and the value $w$ in the form $a+bi$. Use the square root symbol $\sqrt{\ \ }$ where needed to give an exact value for your answer. Be sure to include parentheses where necessary, e.g. to distinguish $\frac{1}{2k}$ from $\frac{1}{2}k$. .
 [/INST] Just do it.   If $z = a + bi$ than $|z| =\sqrt{a^2 + b^2}$ and
$z = \sqrt{a^2+ b^2}(\frac a{a^2 + b^2} + \frac b{a^2 + b^2} i)$.  Let $r= \sqrt{a^2 + b^2} = |z|$.
Now $\frac a{a^2 + b^2}=x$ and $\frac b{a^2 + b^2}=y$ are two numbers such that $x^2 + y^2 = 1$.  That means that there must be some angle $\theta$ so that $x = \cos \theta$ and $y = \sin \theta$.  What angle can that be?
Well, $\tan \theta = \frac {\sin \theta}{\cos \theta} = \frac yx = \frac {\frac b{a^2 + b^2}}{\frac a{a^2 + b^2}}= \frac ba$.  So $\theta = \arctan \frac ba$.
so $z =a+bi = \sqrt{a^2 + b^2}(\cos \arctan \frac ba + \sin \arctan \frac ba i)$.
Now by definition $e^{\psi i} = \cos \psi + i\sin \psi$ so
$z = \sqrt{a^2 + b^2}e^{\arctan \frac ba i}$.
So for $z = a+bi; z\ne 0$ we can always convert it to $z= re^{\theta i}$ form by letting $r = |z| =\sqrt{a^2 + b^2}$ and letting $\theta = \arctan \frac ba$.  Always.
....
And converting $z= re^{\theta i}$ to $a + bi$ form is simply noting:
$z = re^{\theta i} = r(\cos \theta + i \sin\theta) =r\cos\theta + r\sin \theta i$ .
So $a = r\cos \theta$ and $b = r\sin \theta$.
That's all there is to it.
so just do them.
If $z = \frac 52 - \frac {5\sqrt 3}2i$ then $r = \sqrt {(\frac 52)^2 + (\frac {5\sqrt 3}2)^2}=???$ and $\theta = \arctan \frac {\frac {5\sqrt 3}2}{\frac 52}=\arctan \frac {5\sqrt 3}{5}=\arctan \sqrt 3 = ????$ and $z = re^{\theta i}$.  That's all.
And if $w = 2e^{\frac {i5\pi}4}$ then $w = 2\cos (\frac {5\pi}4) +2\sin(\frac {5\pi}4)i$.
That's all...
..... but.... wait a minute.  The problem you posted has them both equal to $0$?????  That makes utterly zero sense.
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