what-is-i-1-

Question Number 5165 by 1771727373 last updated on 24/Apr/16

w h a t i s

\sqrt{i + 1} ?

Commented by FilupSmith last updated on 24/Apr/16

assuming i^{2} = - 1,

according to WolframAlpha :

\sqrt{i + 1} =^{4} \sqrt{2} e^{i π / 8} =^{4} \sqrt{2} \cos (\frac{π}{8}) + i^{4} \sqrt{2} \sin (\frac{π}{8})

I am unsure how to work out the values

^{4} \sqrt{2} and \frac{π}{8}

Answered by 123456 last updated on 24/Apr/16

i + 1 = \frac{\sqrt{2}}{\sqrt{2}} (1 + i)

1 + i = \sqrt{2} (\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}})

1 + i = \sqrt{2} (\cos \frac{π}{4} + i \sin \frac{π}{4})

1 + i = \sqrt{2} e^{i (\frac{π}{4} + 2 π k)} k \in Z

\sqrt{1 + i} = \sqrt{\sqrt{2}} e^{i (\frac{π}{8} + π k)}

\sqrt{1 + i} = \sqrt[4]{2} e^{i (\frac{π}{8} + π k)}

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