I am presented with a problem, which asks you to maximize $|z|$ with the constraint of $$\left|z+\frac 2z \right| = 2.$$
I tried to approach this with components, however this degraded into a messy cubic, letting $z=a+bi$ only to receive $$\frac{a^3+a^2b+ab^2+b^3+2a-2b}{a^2+b^2},$$ trying to maximize $\sqrt{a^2+b^2}.$
How does one go through solving this, without calculus?
Your approach works (It seems that you have an error in your approach).
Letting $z=a+bi$ where $a,b$ are real numbers, we have $$\begin{align}&\bigg|a+bi+\frac 2{a+bi}\bigg|=2 \\\\&\iff \bigg|a+bi+\frac{2(a-bi)}{a^2+b^2}\bigg|=2 \\\\&\iff \bigg|\bigg(a+\frac{2a}{a^2+b^2}\bigg)+\bigg(b-\frac{2b}{a^2+b^2}\bigg)i\bigg|=2 \\\\&\iff \sqrt{\bigg(a+\frac{2a}{a^2+b^2}\bigg)^2+\bigg(b-\frac{2b}{a^2+b^2}\bigg)^2}=2 \\\\&\iff \frac{(a(a^2+b^2)+2a)^2+(b(a^2+b^2)-2b)^2}{(a^2+b^2)^2}=4 \\\\&\iff \frac{(a^2 + b^2) (a^4 + 2 a^2 b^2 + 4 a^2 + b^4 - 4 b^2 + 4)}{(a^2+b^2)^2}=4 \\\\&\iff \frac{a^4 + 2 a^2 b^2 + 4 a^2 + b^4 - 4 b^2 + 4}{a^2 + b^2}=4 \\\\&\iff a^4 + 2 a^2 b^2 + 4 a^2 + b^4 - 4 b^2 + 4-4(a^2+b^2)=0 \\\\&\iff b^4 +(2 a^2 - 8) b^2 +a^4 + 4=0 \\\\&\iff b^2=4-a^2\pm\sqrt{12-8a^2}\end{align}$$
So, we can write $$|z|=\sqrt{a^2+b^2}=\sqrt{4\pm\sqrt{12-8a^2}}$$
So, the maximum value of $|z|$ is $$\sqrt{4+\sqrt{12-8\times 0^2}}=\sqrt{4+2\sqrt 3}=\sqrt{(1+\sqrt 3)^2}=\color{red}{1+\sqrt 3}$$ which is attained when $z=\pm(1+\sqrt 3)i$.
Another approach (there was an answer using this approach, but the answer was deleted) :
For any two complex numbers $z_1$ and $z_2$, $$|z_1+z_2|\le |z_1|+|z_2|$$ (see here for a proof)
Using this inequality, we have $$|z|=\bigg|z+\frac 2z-\frac 2z\bigg|\le \bigg|z+\frac 2z\bigg|+\bigg|-\frac 2z\bigg|=2+\frac{2}{|z|}$$ $$\implies |z|\le 2+\frac{2}{|z|}\implies |z|^2-2|z|-2\le 0\implies |z|\le 1+\sqrt{3}$$
If $z=(1+\sqrt 3)i$, then $|z|=1+\sqrt 3$ and $\bigg|z+\dfrac 2z\bigg|=2$ hold.
Therefore, the maximum value of $|z|$ is $\color{red}{1+\sqrt 3}$.
