i\sqrt{i}?

i=\sqrt{i}=?

#Algebra

Note by Fahim Muhtamim
1 year, 7 months ago

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Comments

Noting first that i=eiπ/2\large i = e^{i \pi/2} and that eix=cos⁡(x)+isin⁡(x)\large e^{ix} = \cos(x) + i \sin(x) we see that

i=(i)1/2=(eiπ/2)1/2=eiπ/4=cos⁡(π/4)+isin⁡(π/4)=12+i×12=1+i2\large \sqrt{i} = (i)^{1/2} = (e^{i \pi/2})^{1/2} = e^{i \pi/4} = \cos(\pi/4) + i\sin(\pi/4) = \dfrac{1}{\sqrt{2}} + i \times \dfrac{1}{\sqrt{2}} = \dfrac{1 + i}{\sqrt{2}}.

Brian Charlesworth - 1 year, 7 months ago

You can assume i=(a+bi)2i=(a+bi)^2 Then, by expanding, we get i=(a2−b2)+2abii=(a^2-b^2)+2abi which means {a2−b2=02ab=1\begin{cases} a^2-b^2=0 \\ 2ab=1 \end{cases} After solving, we get a=b=12a=b=\dfrac{1}{\sqrt{2}} (for a,b>0a,b>0) . Therefore, i=12+i2\sqrt{i}=\dfrac{1}{\sqrt{2}}+\dfrac{i}{\sqrt{2}}

Isaac YIU Math Studio - 1 year, 7 months ago

Brilliant!!

Fahim Muhtamim - 1 year, 7 months ago
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