+0  
 
+20
644
7
avatar+7220 

\(\sqrt{i}=\sqrt{\sqrt{-1}}\)

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

Best Answer 

 #3
avatar+97561 
+35

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

 Jul 12, 2016
 #1
avatar+135 
+30

Hello Max,

you may find this interesting to read

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

 Jul 12, 2016
 #2
avatar
+30

(-1)^(1/4)=

0.70710678118654752440084... +
0.70710678118654752440084... i

 Jul 12, 2016
 #3
avatar+97561 
+35
Best Answer

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

Melody Jul 12, 2016
 #4
avatar+7220 
+15

\(\text{Why is }\sqrt{i}=\frac{1+i}{\sqrt2}? \text{I still don't understand how to obtain this answer?}\)

.
 Jul 16, 2016
 #6
avatar+27476 
+25

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
 #5
avatar
+20

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
 #7
avatar+7220 
+15

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

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