square root of i
- Thread starter helpbaah
- Start date
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,089
It means nothing.
What you presumably mean is that i is at an angle of 90 degrees (from the real axis), so its square root is at half that, 45 degrees. This is because squaring doubles the angle (and squares the modulus). So, to find the square root of i, you can find the complex number with angle 45 degrees and modulus 1.
But what you wrote isn't something we'd ever write. So what do you really want to know? What have you actually seen written that you don't understand? That will be easier to explain, in the form you've learned it.
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
In the complex plane, "i" is directly above the origin at distance 1. In polar coordinates that would be \(\displaystyle r= 1\), \(\displaystyle \theta= \pi/2\). We can write \(\displaystyle i= 1(cos(\pi/2))+ 1(sin(\pi/2)i)\) since \(\displaystyle cos(\pi/2)= 0\) and \(\displaystyle sin(\pi/12)= 1\). Since \(\displaystyle e^{ix}= cos(x)+ isin(x)\), we an also write \(\displaystyle i= e^{i\pi/2}\) so that \(\displaystyle \sqrt{i}= (e^{i\pi/2})^{1/2}= \)\(\displaystyle e^{i\pi/4}= cos(\pi/4)+ i sin(\pi/4)= \)\(\displaystyle \frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2}\).
Of course every number has two square roots. To get the other one, use the fact that sine and cosine are "periodic" with period \(\displaystyle 2\pi\), \(\displaystyle cos(\theta+ 2\pi)= cos(\theta)\) and \(\displaystyle sin(\theta+ 2\pi)= sin(\theta)\). So \(\displaystyle i= e^{i(\pi/2+ 2\pi)}= e^{5\pi/2}\) also. But dividing the exponent by 2 breaks that identity. The other value for \(\displaystyle \sqrt{i}\) is \(\displaystyle \sqrt{i}= \cos\left(\frac{5\pi}{2}\right)+ i\sin\left(\frac{5\pi}{2}\right)= -\frac{\sqrt{2}}{2}- i\frac{\sqrt{2}}{2}\).
It should be no surprise that the two square roots are negatives of each other: \(\displaystyle (-x)^2= x^2\) even for complex numbers.
Of course every number has two square roots. To get the other one, use the fact that sine and cosine are "periodic" with period \(\displaystyle 2\pi\), \(\displaystyle cos(\theta+ 2\pi)= cos(\theta)\) and \(\displaystyle sin(\theta+ 2\pi)= sin(\theta)\). So \(\displaystyle i= e^{i(\pi/2+ 2\pi)}= e^{5\pi/2}\) also. But dividing the exponent by 2 breaks that identity. The other value for \(\displaystyle \sqrt{i}\) is \(\displaystyle \sqrt{i}= \cos\left(\frac{5\pi}{2}\right)+ i\sin\left(\frac{5\pi}{2}\right)= -\frac{\sqrt{2}}{2}- i\frac{\sqrt{2}}{2}\).
It should be no surprise that the two square roots are negatives of each other: \(\displaystyle (-x)^2= x^2\) even for complex numbers.
Steven G
Elite Member
- Joined
- Dec 30, 2014
- Messages
- 14,383
I need to remember that. You say the most amazing things! Is it also true for even and irrational numbers? Hmm, aren't even and irrational numbers part of the complex numbers? I learn so much on this forum!
