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# What is square root of i?

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$$\sqrt{i}=\sqrt{\sqrt{-1}}$$

And are there any other ways to present this special number?

I also knows $$\sqrt{i}=(-1)^{\frac{1}{4}}=e^{\frac{i\pi}{4}}$$

Jul 12, 2016
edited by MaxWong  Jul 12, 2016
edited by MaxWong  Jul 12, 2016

#3
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http://www.wolframalpha.com/input/?i=sqrt(i)

Jul 12, 2016

#1
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Hello Max,

you may find this interesting to read

https://www.math.toronto.edu/mathnet/questionCorner/rootofi.html

Jul 12, 2016
#2
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(-1)^(1/4)=

0.70710678118654752440084... +
0.70710678118654752440084... i

Jul 12, 2016
#3
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http://www.wolframalpha.com/input/?i=sqrt(i)

Melody Jul 12, 2016
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$$\text{Why is }\sqrt{i}=\frac{1+i}{\sqrt2}? \text{I still don't understand how to obtain this answer?}$$

.
Jul 16, 2016
#6
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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:  $$\sqrt{i}=\frac{1+i}{\sqrt{2}}$$

.

Alan  Jul 16, 2016
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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,

Jul 16, 2016
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Guest #5, I don't quite understand your answer. But Alan's answer did help me a lot. Thanks Alan. And Guest#5 can you give a more detailed explanation?

MaxWong  Jul 16, 2016