Leysa sqrt{p}=hsqrt{R}

$$\sqrt { p } = h \sqrt { R }$$

Leystu fyrir R (complex solution)

$$\left\{\begin{matrix}R=\frac{p}{h^{2}}\text{, }&h\neq 0\text{ and }\left(|-\frac{arg(p)}{2}+arg(\sqrt{\frac{p}{h^{2}}}h)|<\pi \text{ or }p=0\right)\\R\in \mathrm{C}\text{, }&p=0\text{ and }h=0\end{matrix}\right.$$

Leystu fyrir h (complex solution)

$$\left\{\begin{matrix}h=R^{-\frac{1}{2}}\sqrt{p}\text{, }&R\neq 0\\h\in \mathrm{C}\text{, }&p=0\text{ and }R=0\end{matrix}\right.$$

Leystu fyrir R

$$\left\{\begin{matrix}R=\frac{p}{h^{2}}\text{, }&\left(p\geq 0\text{ and }h>0\right)\text{ or }\left(h\neq 0\text{ and }p=0\right)\\R\geq 0\text{, }&h=0\text{ and }p=0\end{matrix}\right.$$

Leystu fyrir h

$$\left\{\begin{matrix}h=\sqrt{\frac{p}{R}}\text{, }&p\geq 0\text{ and }R>0\\h\in \mathrm{R}\text{, }&p=0\text{ and }R=0\end{matrix}\right.$$