What is the square root of imaginary numbers?

So -1 has a square root of i. But what is the square root of i?What is the square root of imaginary numbers?
When it gets confusing, just google it. Google has a built in calculator that is better than the Windows Calculator. You just have to learn the codes...





Simply input: ';sqrt i'; into the search box.





If you have a TI Graphing Calculator you can also use the square root button and the i key. However you will have to turn on imaginary numbers through the mode screen.What is the square root of imaginary numbers?
The question doesn't make sense, and the answer is also meaningless.


Yes, i is the square root of -1.


The suqare root of i? What do you want, another letter? b? r? w? n?


That's like saying ';2+2=4, but what does 4+4 equal?'; Report Abuse

hi


Do you know Euhler's formula?





According to that formulae, any complex no can be written as





z=r x e^ia





where a is the argument of a complex no, and r is the magnitude of a complex number.





So, square root of z will be sqrt(r) x e^i(a/2)





here e^i(a/2) can be written cos(a/2)+isin(a/2)





so i can be written as e^i(pi/2)





then its square root is e^i(pi/4) .





bye


:)
The square root of 'i' cannot be simplified when stated as an algebraic expression.





That is to say your ';square root of i'; is nothing more than the square root of i.





@All the people who answer with ';Euhler's (sic)'; number and pi:





WELL LET ME SEE...





Any number 'x' can be represented in the form


(ciel((2x^17 + 2x^3 / 4)(2) + 2abs((ln(e^2)^(i*sqrt(6sum(n=0, INF, (-1)^n/(2n+1)))))))+1-1+1-1+1-1+1-1)*0+x





So the answer must be





(ciel((2sqrt(i)^17 + 2root(-1, 4)^3 / 4)(2) + 2abs((ln(e^2)^(i*sqrt(6sum(n=0, INF, (-1)^n/(2n+1)))))))+1-1+1-1+1-1+1-1)*0-r鈥?633825300114114700748351602688)
Let x = (i)^1/2





Therefore x = (0 + i)^(1/2)





As every complex number can represented as r(cos (a) + i Sin (a))





Here r = 1, a = (Pi / 2)





This can be written as x = 1(cos(Pi/2) + i sin (Pi/2))^ (1/2)





OR x = (cos (2n(Pi) + (Pi/2)) + i sin (2n(Pi) + (Pi/2)))^(1/2)





By De' Moiver's Theorem r(cos (a) + i sin (a))^ n = r(cos (na) + i sin(na)) = cis (na)





therefore x = cis (Pi (4n + 1)/4)





where n = 0 , 1





Therefore the roots are:





1) cis (Pi/ 4)


2) cis (5(Pi)/4)
The answer is whatever you imagine it to be when dealing with things imaginary.





Thank you for wasting our time!
The square root of i is the fourth root of -1.

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