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Question Number 139608 by mohammad17 last updated on 29/Apr/21

find the first root (−8i)^(1/2)

$${find}\:{the}\:{first}\:{root}\:\left(−\mathrm{8}{i}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$

Answered by floor(10²Eta[1]) last updated on 29/Apr/21

(√(−8i))=a+bi −8i=a^2 −b^2 +2abi ⇒a^2 −b^2 =0∧−8=2ab ∣a∣=∣b∣∧ab=−4 b^2 =4⇒b=±2, a=±2 (√(−8i))=2−2i or −2+2i

$$\sqrt{−\mathrm{8i}}=\mathrm{a}+\mathrm{bi} \\ $$$$−\mathrm{8i}=\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} +\mathrm{2abi} \\ $$$$\Rightarrow\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} =\mathrm{0}\wedge−\mathrm{8}=\mathrm{2ab} \\ $$$$\mid\mathrm{a}\mid=\mid\mathrm{b}\mid\wedge\mathrm{ab}=−\mathrm{4} \\ $$$$\mathrm{b}^{\mathrm{2}} =\mathrm{4}\Rightarrow\mathrm{b}=\pm\mathrm{2},\:\mathrm{a}=\pm\mathrm{2} \\ $$$$\sqrt{−\mathrm{8i}}=\mathrm{2}−\mathrm{2i}\:\mathrm{or}\:−\mathrm{2}+\mathrm{2i} \\ $$

Answered by mr W last updated on 29/Apr/21

−8i=8e^((2kπ−(π/2))i) (−8i)^(1/2) =2(√2)e^((k−(1/4))πi) ,k=0,1 =2(√2)((1/( (√2)))−(1/( (√2)))i)=2−2i or =2(√2)(−(1/( (√2)))+(1/( (√2)))i)=−2+2i

$$−\mathrm{8}{i}=\mathrm{8}{e}^{\left(\mathrm{2}{k}\pi−\frac{\pi}{\mathrm{2}}\right){i}} \\ $$$$\left(−\mathrm{8}{i}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{2}\sqrt{\mathrm{2}}{e}^{\left({k}−\frac{\mathrm{1}}{\mathrm{4}}\right)\pi{i}} ,{k}=\mathrm{0},\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\sqrt{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{i}\right)=\mathrm{2}−\mathrm{2}{i} \\ $$$$\:{or}\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\sqrt{\mathrm{2}}\left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{i}\right)=−\mathrm{2}+\mathrm{2}{i} \\ $$

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