Question Number 130639 by mohammad17 last updated on 27/Jan/21
$${for}\:{any}\:{two}\:{complex}\:{number}\:\left({z},{w}\right)\:{then} \\ $$$$\sqrt{{z}.{w}}=\sqrt{{z}}.\sqrt{{w}} \\ $$$$ \\ $$$${true}\:\:{or}\:{false} \\ $$
Commented by malwan last updated on 27/Jan/21
$${thank}\:{you} \\ $$
Commented by mohammad17 last updated on 27/Jan/21
$${can}\:{you}\:{give}\:{me}\:{example}\:{sir} \\ $$
Commented by mohammad17 last updated on 27/Jan/21
$${is}\:{true}\:{sir}\:{not}\:{false} \\ $$
Answered by MJS_new last updated on 27/Jan/21
$${z}={r}\mathrm{e}^{\mathrm{i}\alpha} ;\:{r}\geqslant\mathrm{0}\wedge\mathrm{0}\leqslant\alpha<\mathrm{2}\pi \\ $$$${w}={s}\mathrm{e}^{\mathrm{i}\beta} ;\:{s}\geqslant\mathrm{0}\wedge\mathrm{0}\leqslant\beta<\mathrm{2}\pi \\ $$$$\sqrt{{z}}=\sqrt{{r}}\mathrm{e}^{\mathrm{i}\frac{\alpha}{\mathrm{2}}} \\ $$$$\sqrt{{w}}=\sqrt{{s}}\mathrm{e}^{\mathrm{i}\frac{\beta}{\mathrm{2}}} \\ $$$$\sqrt{{z}}\sqrt{{w}}=\sqrt{{rs}}\mathrm{e}^{\mathrm{i}\frac{\alpha+\beta}{\mathrm{2}}} \\ $$$$\sqrt{{zw}}=\sqrt{{r}\mathrm{e}^{\mathrm{i}\alpha} {s}\mathrm{e}^{\mathrm{i}\beta} }=\sqrt{{rs}\mathrm{e}^{\left(\alpha+\beta\right)} }=\sqrt{{rs}}\mathrm{e}^{\mathrm{i}\frac{\alpha+\beta}{\mathrm{2}}} \\ $$$$\Rightarrow\:\mathrm{true} \\ $$
Commented by mohammad17 last updated on 27/Jan/21
$${thank}\:{you}\:{sir} \\ $$