Express in regular form the conjugates of $z \in c$ that satisfy $z^2 + 4i = 0$
Let $$ z=re^{i\theta} $$ Thus $z^2 = -4i = 4e^{i\frac{3\pi}2} $
$r^2=4, r=2$
$\theta_1 = \frac{3\pi}4, \theta_2 = \frac{7\pi}4 $
Thus, $z_1 =2e^{i\frac{3\pi}4}, z_2 =2e^{i\frac{7\pi}4} $
Here comes the part I don't get it. How did they get the rectangular from the exponential or polar form?
$z_1 = -\sqrt2 +\sqrt2i, z_2 = \sqrt2 -\sqrt2i $
$\bar{z_1} = -\sqrt2 -\sqrt2i, \bar{z_2} =\sqrt2 +\sqrt2i $
The complex conjugate of the polar form of a complex number is given by $$\overline{re^{i\theta}}=re^{-i\theta}.$$
To see why, use Euler's formula: \begin{align} \overline{r(\cos\theta+i\sin\theta)}&=\overline{r}\ \overline{(\cos\theta+i\sin\theta)}\\ &=r\,(\cos\theta-i\sin\theta)\\ &=r\,(\cos(-\theta)+i\sin(-\theta))\quad\text{[since cosine is even and sine is odd]}\\ &=r\,e^{-i\theta}. \end{align}
So you can find the conjugates of your $z$ without converting first to rectangular form.
Of course, if you do want to convert from polar to rectangular form, just use Euler's formula employed above:
$$z=re^{i\theta}=r(\cos\theta+i\sin\theta)$$
which can be converted to rectangular form $z=a+bi$ via $$a=r\cos\theta,\quad b=r\sin\theta,$$ which in your case leads to $a=2\cos(3\pi/4)=-\sqrt 2$ and $b=2\sin(3\pi/4)=\sqrt 2$. So $$z=2e^{i{3\pi\over 4}}=-\sqrt{2}+i\sqrt{2}\implies \overline{z}=2e^{-i{3\pi\over 4}}=-\sqrt{2}-i\sqrt{2}.$$ You can do the other one.
