Find both square roots of a complex number

Let a square root be $a+bi$, where $a,b\in\mathbb R$. Then

$\begin{aligned}15+8i&=(a+bi)^2\\&=a^2+2abi+b^2i^2\\&=a^2+2abi-b^2.\end{aligned}$

Comparing components gives $a^2-b^2=15$ (1) and $2ab=8$, hence $a=4/b$ (2). The division is valid because $ab=4$ forces $b\ne0$. Substitute (2) into (1):

$\begin{aligned}\left(\frac4b\right)^2-b^2&=15,\\16-b^4&=15b^2,\\b^4+15b^2-16&=0,\\(b^2+16)(b^2-1)&=0.\end{aligned}$

Reject $b^2=-16$ because $b$ is real. Thus $b^2=1$ and $b=\pm1$. When $b=1$, $a=4$; when $b=-1$, $a=-4$. The two roots are $4+i$ and $-4-i$. The signs must stay paired.

On the calculator shown in the notes, press MODE and highlight a+bi, so results use algebraic form $x+iy$. Enter $\sqrt{15+8i}$: the screen shows $4+i$. Then enter $-\sqrt{15+8i}$: the screen shows $-4-i$.

The square-root command returns one selected root. Multiply that result by $-1$ to obtain the other root; one calculator output is not the complete answer to a request for both roots.