+136 Hello Max,
you may find this interesting to read
https://www.math.toronto.edu/mathnet/questionCorner/rootofi.html
+9673 \(\text{Why is }\sqrt{i}=\frac{1+i}{\sqrt2}? \text{I still don't understand how to obtain this answer?}\)
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+33661 In general: \(e^{i\theta}=\cos \theta+i\sin\theta\)
You already know that \(\sqrt{i}=e^{i\pi /4}\)
Put these together to get: \(\sqrt{i}=\cos{(\pi /4)+i\sin{(\pi /4)}}\)
Since we have: \(\cos{(\pi /4)}=\frac{1}{\sqrt{2}} \space \text{and }\sin{(\pi /4)}=\frac{1}{\sqrt{2}}\)
we can see that: \(\sqrt{i}=\frac{1+i}{\sqrt{2}}\)
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Here is a direct algebraic answer:
Suppose that z=c+di, and we want to find √z=a+bi lying in the first two quadrants. So what are a and b?
Precisely we have:
a=Sqrt[c+sqrt(c^2+d^2)/2]
and
b=d/|d|Sqrt[-c+sqrt(c^2+d^2)/2]
(The factor of d/|d| is used so that b has the same sign as d) To find this, we can use brute force and the quadratic formula. Squaring, we would need to solve:
a^2−b^2+2abi=c+di.
This gives two equations and two unknowns (seperate into real and imaginary parts) which can then be solved by substitutions and the quadratic formula.
Hope that helps,
