Why is the following equivalent transformation of the imaginary number legitimate?

Question

Why is this substitution acceptable?

$\sqrt{i}=\frac{1+i}{\sqrt{2}}$

See also here: http://math.stackexchange.com/questions/44406/how-do-i-get-the-square-root-of-a-complex-number — Hans Lundmark, Oct 26 '14 at 12:06

score 2 · Accepted Answer · answered Oct 26 '14 at 11:36

2

Because $i = e^{i\frac{\pi}{2}}$ and $\sqrt{i} = i^{\frac{1}{2}} = e^{i\frac{\pi}{4}} = \cos(\frac{\pi}{4}) + i \cdot \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i$.

answered Oct 26 '14 at 11:36

GDumphart

2,250

score 1 · Answer 2 · answered Oct 26 '14 at 11:36

1

$$a+ib=re^{i\theta}$$ where $r^2=|a^2+b^2|$ & $ \theta=\tan^{-1}\left(\frac{b}{a}\right)$

Then $$i=e^{i\pi/2}$$

$$i^{1/2}=e^{i\pi/4}=\frac{1}{\sqrt2}+i\frac{1}{\sqrt 2}=\frac{1+i}{\sqrt 2}$$

answered Oct 26 '14 at 11:36

score 0 · Answer 3 · answered Oct 26 '14 at 12:14

Suppose that $$\sqrt{i} = a+bi$$ Then, as per Yves Daoust's suggestion, $$\sqrt{i}^2=i=0+i=(a+bi)^2=a^2+2abi+b^2i^2=(a^2-b^2)+(2ab)i$$ summarizing we have $$0+i=(a^2-b^2)+(2ab)i$$ All that is now required is to solve the system $$0=a^2+b^2$$ $$1=2ab$$

Why is the following equivalent transformation of the imaginary number legitimate?

3 Answers3