Difficulties with complex numbers and complex planes 
Find the general form of a complex number $z$ such that:
  i) $z^2$ is a real number (i.e. $\operatorname{Im} z^2 = 0$);
  ii) $z^2$ is a purely imaginary number (i.e. $\operatorname{Re} z^2 = 0$);
  iii) $z^2 = \bar z$.

I'm not sure how to tackle this question
For i), I wrote
$$\operatorname{Im}(z^2) = (z^2-z'^2)/2$$
and I then get stuck and I don't know how to rearrange the formula in a helpful way.
 A: Let $z=x+iy$.  Then, $z^2=(x^2-y^2)+i2xy$.  So, if the imaginary part is zero, then 
$$xy=0\implies \bbox[5px,border:2px solid #C0A000]{x=0 \,\,\text{or}\,\, y=0}$$
If the real part of $z^2$ is zero, then 
$$x^2-y^2=0\implies \bbox[5px,border:2px solid #C0A000]{y=\pm x}$$
And finally, if $z^2=\bar z$, then
$$x^2-y^2=x\,\,\text{and}\,\,2xy=-y\implies\bbox[5px,border:2px solid #C0A000]{ (x,y)=(0,0)\,,(1,0)\,,(-1/2,\sqrt{3}/2)\,,(-1/2,-\sqrt{3}/2)}$$
A: Hint:
use the polar form $z=\rho e^{i\theta}$ and your problem becomes simpler since $z^2=\rho^2 e^{i2\theta}$ ....
A: In addition to thinking about this algebraically, you can think about it geometrically.
Do you know about the Argand Plane? This is a way of thinking about complex numbers as points on a 2D plane. (Or as elements of $\mathbb{R}^2$, if that's familiar to you). The $x$-axis is the real axis and the $y$-axis is the imaginary axis. Each point in 2D space (and each complex number) makes a certain angle $\theta$ with the $x$-axis and has a certain magnitude $r$. The angle is called the argument and the magnitude is called the phase, don't ask me why.

(Picture stolen from my link). That's all it takes to uniquely specify a complex number. Here's a question that's fun and important -- given a complex number in $x + iy$ form, how would you find its phase and its argument?
But actually you don't need to know that to answer these problems. When you multiply two complex numbers, you add their angles and multiply their magnitudes.
For example, if you multiply $i \times i$, the result has magnitude $1 \times 1 = 1$ and angle $90° + 90° = 180°$. The answer is the number that has angle $180°$ and lies on the unit circle, and that number is $-1$.
So to answer your first question, we just have to think "which numbers, if I rotate them by themselves, will land on the real axis?" Clearly $1$ and all positive real numbers will do this (their angle is $0°$, so they don't go anywhere when rotated). Also $-1$ and all negative real numbers, since $180° + 180° = 360° = 0°$. With a bit more thought, we can see that $i$ and all positive imaginary numbers will land on the negative axis when squared, since $90° + 90° = 180°$. And with a bit more, all negative imaginary numbers, since $270° + 270° = 540° = 180°$. Personally I think about algebra when I think about $i \times i = -1$ and $-i \times -i = -1$, but it can be enlightening to see how it works out in this way as well.
For your second question, we're looking for angles that add to $90°$ and $270°$. So we've got $45°$ and $135°$, but also $225° + 225° = 90°$ and $315° + 315° = 270°$. I got those by adding $45°$ to $180°$ and to $270°$.
For your last question, the complex conjugate of a number should just be its reflection across the $x$-axis (since you're negating the imaginary part). So you're looking for which rotates itself to its own reflection.
