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Question Number 136154 by sahiljakhar last updated on 19/Mar/21

Answered by sahiljakhar last updated on 19/Mar/21

$$\sqrt{z}=\frac{z}{x}$$

Commented by mathmax by abdo last updated on 19/Mar/21

$$\mathrm{i}\mathrm{suppose}\mathrm{x}\mathrm{given}\mathrm{and}\mathrm{z}\mathrm{unknown}\Rightarrow \frac{\sqrt{\mathrm{z}}}{\mathrm{z}}=\frac{1}{\mathrm{x}}(\mathrm{x}\ne 0)$$$$\mathrm{let}\mathrm{z}={\mathrm{re}}^{\mathrm{i}\theta }\Rightarrow \frac{\sqrt{\mathrm{r}}{\mathrm{e}}^{\frac{\mathrm{i}\theta }{2}}}{\mathrm{r}{\mathrm{e}}^{\mathrm{i}\theta }}=\frac{1}{\mathrm{x}}\Rightarrow \frac{1}{\sqrt{\mathrm{r}}}{\mathrm{e}}^{\frac{\mathrm{i}\theta }{2}-\mathrm{i}\theta }=\frac{1}{\mathrm{x}}\Rightarrow {\mathrm{e}}^{\mathrm{i}(-\frac{\theta }{2})}=\frac{\sqrt{\mathrm{r}}}{\mathrm{x}}\Rightarrow$$$${\mathrm{e}}^{\frac{\mathrm{i}\theta }{2}}=\frac{\mathrm{x}}{\sqrt{\mathrm{r}}}\mathrm{if}\mid \frac{\mathrm{x}}{\sqrt{\mathrm{r}}}\mid >1\Rightarrow \mathrm{no}\mathrm{solutin}\mathrm{if}\mid \frac{\mathrm{x}}{\sqrt{\mathrm{r}}}\mid <1\mathrm{let}\frac{\mathrm{x}}{\sqrt{\mathrm{r}}}={\mathrm{e}}^{\mathrm{ia}}$$$$\mathrm{e}\Rightarrow \frac{\theta }{2}=\mathrm{a}+2\mathrm{k}\pi \Rightarrow \theta =2\mathrm{a}+4\mathrm{k}\pi (\mathrm{k}\in \mathrm{Z})$$$$\mathrm{if}\mathrm{x}\mathrm{real}\Rightarrow \cos \left(\frac{\theta }{2}\right)+\mathrm{isin}\left(\frac{\theta }{2}\right)=\frac{\mathrm{x}}{\sqrt{\mathrm{r}}}\Rightarrow \sin \left(\frac{\theta }{2}\right)=0\Rightarrow \theta =2\mathrm{k}\pi$$

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