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Question Number 5165 by 1771727373 last updated on 24/Apr/16
what is  (√(i+1))    ?
$${what}\:{is} \\ $$$$\sqrt{{i}+\mathrm{1}}\:\:\:\:? \\ $$
Commented by FilupSmith last updated on 24/Apr/16
assuming i^2 =−1,  according to WolframAlpha:  (√(i+1))=^4 (√2)e^(iπ/8) =^4 (√2)cos((π/8))+i^4 (√2)sin((π/8))    I am unsure how to work out the values  ^4 (√2)   and  (π/8)
$$\mathrm{assuming}\:{i}^{\mathrm{2}} =−\mathrm{1}, \\ $$$$\mathrm{according}\:\mathrm{to}\:\mathrm{WolframAlpha}: \\ $$$$\sqrt{{i}+\mathrm{1}}=\:^{\mathrm{4}} \sqrt{\mathrm{2}}{e}^{{i}\pi/\mathrm{8}} =\:^{\mathrm{4}} \sqrt{\mathrm{2}}\mathrm{cos}\left(\frac{\pi}{\mathrm{8}}\right)+{i}\:^{\mathrm{4}} \sqrt{\mathrm{2}}\mathrm{sin}\left(\frac{\pi}{\mathrm{8}}\right) \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{am}\:\mathrm{unsure}\:\mathrm{how}\:\mathrm{to}\:\mathrm{work}\:\mathrm{out}\:\mathrm{the}\:\mathrm{values} \\ $$$$\:^{\mathrm{4}} \sqrt{\mathrm{2}}\:\:\:\mathrm{and}\:\:\frac{\pi}{\mathrm{8}} \\ $$
Answered by 123456 last updated on 24/Apr/16
i+1=((√2)/( (√2)))(1+i)  1+i=(√2)((1/( (√2)))+i(1/( (√2))))  1+i=(√2)(cos (π/4)+isin (π/4))  1+i=(√2)e^(i((π/4)+2πk))             k∈Z  (√(1+i))=(√(√2))e^(i((π/8)+πk))   (√(1+i))=(2)^(1/4) e^(i((π/8)+πk))
$${i}+\mathrm{1}=\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{1}+{i}\right) \\ $$$$\mathrm{1}+{i}=\sqrt{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+{i}\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$\mathrm{1}+{i}=\sqrt{\mathrm{2}}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{4}}+{i}\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\right) \\ $$$$\mathrm{1}+{i}=\sqrt{\mathrm{2}}{e}^{{i}\left(\frac{\pi}{\mathrm{4}}+\mathrm{2}\pi{k}\right)} \:\:\:\:\:\:\:\:\:\:\:\:{k}\in\mathbb{Z} \\ $$$$\sqrt{\mathrm{1}+{i}}=\sqrt{\sqrt{\mathrm{2}}}{e}^{{i}\left(\frac{\pi}{\mathrm{8}}+\pi{k}\right)} \\ $$$$\sqrt{\mathrm{1}+{i}}=\sqrt[{\mathrm{4}}]{\mathrm{2}}{e}^{{i}\left(\frac{\pi}{\mathrm{8}}+\pi{k}\right)} \\ $$

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