Why does ${\sqrt{\pm i} = 1/\sqrt{2}\pm i/\sqrt{2}}$?

Question

So I'm given the equation ${y^{(4)}+y =0}$. My goal is to find the general solution. I got to the point where ${r=\sqrt{\pm i}}$ and when looking at the solution, they say that that ${\sqrt{\pm i} = 1/\sqrt{2}\pm i/\sqrt{2}}$. I'm trying to understand how that relationship makes sense, but can't seem to come up with an answer. Any help would be greatly appreciated.

Answer 1

Start with Euler's identity: $$e^{i\pi}=-1\\=i^2$$ Therefore $$e^{i\pi/2}=i$$ Which gives $$i^{1/2}=e^{i\pi/4}$$ Recall Euler's formula: $$e^{ix}=\cos x+i\sin x$$ $x=\pi/4$ gives $$i^{1/2}=\cos(\pi/4)+i\sin(\pi/4)$$ $$i^{1/2}=2^{-1/2}(1+i)$$ Note that $(-i)^{1/2}=(-1)^{1/2}i^{1/2}=i\cdot i^{1/2}$ Which you can figure out.

Answer 2

The square root of a complex number can be defined as $e^{\frac12 \log(z)}$, where $\log(z)=\log|z|+i\arg(z)$ is defined on the principal branch $(-\pi,\pi)$.

Well, $e^{\frac12 \log(i)}=e^{\frac{i\pi}4}=\cos(\frac\pi4)+i\sin(\frac\pi4)=\frac1{\sqrt{2}}+i\frac1{\sqrt{2}}$. Can you see why $e^{\frac12\log(-i)}=\frac1{\sqrt{2}}-i\frac1{\sqrt{2}}$?

You need some knowledge on complex analysis to justify this, namely what is really the complex logarithm function and Euler's formula.

Answer 3

You need to solve $\lambda^4+1=0$. Complete the square in a slightly unconventional way by supplementing the middle term in the binomial formula, and apply the difference-of-squares formula $$ 0=(λ^2+1)^2-2λ^2=(λ^2+\sqrt{2}λ+1)(λ^2-\sqrt{2}λ+1) $$ Both factors are now quadratic polynomials with real coefficients, so that the standard solution method applies.

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As to the computation of the square roots, there you want to solve $-i=(a+ib)^2=(a^2-b^2)+2iab$ so that solutions have $|a|=|b|$ and $2ab=-2$, thus $a=-b=\pm\frac{\sqrt2}2$.