Argument for $(a+bi)^2$

Question

I found out the modulus for $(a+bi)^2$, which is $$a^2+b^2$$ but I am unable to find the argument.

I found out that $$\theta = \frac{2ab}{(a-b)(a+b)}$$

I don't know how to simplify further! Please help!!!!

Answer 1

The argument of $(a+bi)^2$ is twice the argument of $a+bi$, since $(re^{i\theta})^2=r^2e^{2i\theta}$.

Answer 2

we can simplify the given complex number as follows $$\left(a+ib\right)^2=(a^2-b^2)+2iab$$ the modulus is $$=\sqrt{(a^2-b^2)^2+(2ab)^2}=\sqrt{(a^2-b^2)^2+4a^2b^2}=\sqrt{a^4+b^4+2a^2b^2}=a^2+b^2$$ & the module is given as

Case 1: $(a+ib)^2$ lies in the I-quadrant i.e. $(a^2-b^2)\geq0$ & $ab\geq0$ then argument is $$\theta=tan^{-1}\left|\frac{4ab}{a^2-b^2}\right|$$ Case 2: $(a+ib)^2$ lies in the II-quadrant i.e. $(a^2-b^2)\leq0$ & $ab\geq0$ then argument is $$\theta=\pi-tan^{-1}\left|\frac{4ab}{a^2-b^2}\right|$$ Case 3: $(a+ib)^2$ lies in the III-quadrant i.e. $(a^2-b^2)\leq0$ & $ab\leq0$ then argument is $$\theta=-\left(\pi-tan^{-1}\left|\frac{4ab}{a^2-b^2}\right|\right)$$ Case 4: $(a+ib)^2$ lies in the IV-quadrant i.e. $(a^2-b^2)\geq0$ & $ab\leq0$ then argument is $$\theta=-tan^{-1}\left|\frac{4ab}{a^2-b^2}\right|$$