A Rather "Complex" Probability

We know that $z^2={\left(x+iy\right)}^2=x^2-y^2+2xyi$ .

Now, for the real part, $|x^2-y^2|\le 1$ for all $|x|\le 1,|y|\le 1$ . Thus, we just have to consider the imaginary part $|2xy|\le 1$ , or $|y|\le \frac{1}{2|x|}$ .

This produces a graph like this . We just need to find the proportion of the square in the red shaded area. By symmetry, we just need to look at the first quadrant.

In the first quadrant, the ratio of the shaded red area to the square is given by $\frac{1}{2}+\int_{\frac{1}{2}}^{1}\frac{dx}{2x}=\frac{1}{2}+\frac{\ln\left(2\right)}{2}=\frac{1+\ln\left(2\right)}{2}$

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For the first bonus question, $z^{-1}={\left(x+iy\right)}^{-1}=\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}$ . Looking at $|\frac{x}{x^2+y^2}|\le 1$ and $|\frac{y}{x^2+y^2}|\le 1$ , you will notice that they form four overlapping circles of radius $\frac{1}{2}$ centered at $(\frac{1}{2},0)$ , $(-\frac{1}{2},0)$ , $(0,\frac{1}{2})$ , and $(0,-\frac{1}{2})$ . With a little geometry, the area this shape total up to $\frac{\pi}{2}+1$ . Divide this by the area of the square to get $\frac{\pi+2}{8}$ .

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Now, for the second bonus question, consider $\displaystyle\lim_{n\to\infty}z^n$ . This will converge if and only if $|z|\le 1$ , or $\sqrt{x^2+y^2}\le 1$ . It also happens that this is the case when $\displaystyle\lim_{n\to\infty}z^n\in S$ . This produces a circle centered at the origin with radius $1$ , having an area of $\pi$ . Divide this by the area of the square to get $\frac{\pi}{4}$ for the probability that $\displaystyle\lim_{n\to\infty}z^n\in S$ .

Simply solve it by geometric probability by considering
-1<= 2xy <=1
This is so because, (x^2 - y^2) is always between -1 and 1 for all x and y belonging to (-1,1).......!!