What's my error ? $i^1$ means rotation of $90°$ in anticlockwise manner from positive real axis. So $i^{1/2}$ means rotation of $45°$. So square root of $i$ must have both part positive (real and imaginary). But answers in all books contain negative answers also. Please guide
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There are two square roots of $i$. – Zain Patel Jul 08 '16 at 08:46
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Rotating $45^{\circ}$ (anticlockwise) twice amounts to a $90^{\circ}$ rotation, but so does rotating by $135^{\deg}$ clockwise (or, if you like $-135^{\circ}$ anticlockwise). – Travis Willse Jul 08 '16 at 08:46
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It is a matter of branch of log: write $i=e^{i\pi/2}=e^{i\pi/2+2\pi i}$, now take the square root and compare the two resulting angles. – b00n heT Jul 08 '16 at 08:47
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$$i^.5=−1^.25=0.707106781+0.707106781i$$ – Amruth A Jul 08 '16 at 09:22
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So your book does not cover De Moivre's formula? – Ziyuan Jul 09 '16 at 11:46
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Why not rotate of 135° clockwise? If we do this twice, the result is the rotation of 90° anticlockwise, right? – Did Jul 09 '16 at 11:54
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Rotating anti-clockwise by 90 degrees corresponds to multiplication by $i$.
Rotating anti-clockwise by 450 degrees is identical to a 90-degree rotation, so it also corresponds to multiplication by $i$.
Taking the square root corresponds to halving the rotation. This gives us 45 degrees (the solution you have found) and 225 degrees (the negative solution in the books).
You might wonder about taking it further: 810, 1170, and so on. This won't give you any new results, because they will each be a multiple of 360 more than the results you've already got.
It is worth getting the picture of multiplication by $i$ corresponding not just to one unique rotation but to an infinite list of rotations 360 degrees apart. Later on, you will easily be able to see - for example - why a number has 3 cube roots (120 degrees apart), 4 fourth roots (90 degrees apart) and so on.
Martin Kochanski
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Observe that if $(a+bi)^2 = i$ for $a,b >0$, then $(-a-bi)^2=i$ also holds.
In general, if $w$ is a square root of a complex number $z$, then $-w$ is also a square root of $z$.
ThePortakal
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I hope that this particular use of binomial theorem with negative root is limited only to square roots, as the roots are different for cube roots, and so on. Please, consider me as a new entrant. – jiten Jan 10 '18 at 15:13