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Question Number 130639 by mohammad17 last updated on 27/Jan/21

$$foranytwocomplexnumber(z,w)then$$$$\sqrt{z.w}=\sqrt{z}.\sqrt{w}$$$$$$$$trueorfalse$$

Commented by malwan last updated on 27/Jan/21

$$thankyou$$

Commented by mohammad17 last updated on 27/Jan/21

$$canyougivemeexamplesir$$

Commented by mohammad17 last updated on 27/Jan/21

$$istruesirnotfalse$$

Answered by MJS_new last updated on 27/Jan/21

$$z=r{\mathrm{e}}^{\mathrm{i}\alpha };r⩾0\wedge 0⩽\alpha <2\pi$$$$w=s{\mathrm{e}}^{\mathrm{i}\beta };s⩾0\wedge 0⩽\beta <2\pi$$$$\sqrt{z}=\sqrt{r}{\mathrm{e}}^{\mathrm{i}\frac{\alpha }{2}}$$$$\sqrt{w}=\sqrt{s}{\mathrm{e}}^{\mathrm{i}\frac{\beta }{2}}$$$$\sqrt{z}\sqrt{w}=\sqrt{rs}{\mathrm{e}}^{\mathrm{i}\frac{\alpha +\beta }{2}}$$$$\sqrt{zw}=\sqrt{r{\mathrm{e}}^{\mathrm{i}\alpha }s{\mathrm{e}}^{\mathrm{i}\beta }}=\sqrt{rs{\mathrm{e}}^{(\alpha +\beta )}}=\sqrt{rs}{\mathrm{e}}^{\mathrm{i}\frac{\alpha +\beta }{2}}$$$$\Rightarrow \mathrm{true}$$

Commented by mohammad17 last updated on 27/Jan/21

$$thankyousir$$

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