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

Answered by sahiljakhar last updated on 19/Mar/21

(√z) = (z/x)

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

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

i suppose x given and z unknown ⇒((√z)/z)=(1/x) (x≠0) let z =re^(iθ) ⇒(((√r)e^((iθ)/2) )/(r e^(iθ) ))=(1/x) ⇒(1/( (√r)))e^(((iθ)/2)−iθ) =(1/x) ⇒e^(i(−(θ/2))) =((√r)/x) ⇒ e^((iθ)/2) =(x/( (√r))) if ∣(x/( (√r)))∣>1 ⇒no solutin if ∣(x/( (√r)))∣<1 let (x/( (√r)))=e^(ia) e ⇒(θ/2)=a+2kπ ⇒θ =2a+4kπ (k ∈Z) if x real ⇒cos((θ/2))+isin((θ/2))=(x/( (√r))) ⇒sin((θ/2))=0 ⇒θ =2kπ

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

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