Square Root of iDate: 03/08/99 at 21:21:00 From: Nadine Schiavo Subject: Square Root of i. In my Honors Algebra II/Trigonometry class, we just completed a section on complex numbers. One of my students asked me the following: The square root of -1 is i, but what is the square root of i? Can you help? Date: 03/09/99 at 08:31:37 From: Doctor Rick Subject: Re: Square Root of i. There are two complex numbers which, when squared, equal i. (The same holds true for any number. But for real numbers, we arbitrarily say that the positive root is THE square root. When the roots have imaginary parts, any such choice would be even more arbitrary, and we do not bother to choose one.) The square roots of i are (1 + i) * sqrt(2)/2 and -(1 + i) * sqrt(2)/2 You can prove this just by squaring each number. To find them in the first place, you can use Euler's formula e^(i*t) = Cos(t)+i*Sin(t) If t = pi/2, you get e^(i*pi/2) = cos(pi/2)+i*sin(pi/2) = 0 + i*1 = i The square root of this is (e^(i*pi/2))^(1/2) = e^(i*pi/4) and using Euler's formula again, we have e^(i*pi/4) = 1/2 + i*1/2 If you set t = 5*pi/2, you again get e^(i*5pi/2) = i. Taking the square root of this and using Euler's formula, you get the other root of i. Euler's formula makes it easy to find powers and roots by working in polar coordinates in the complex plane. Any number x + iy can be written in terms of a radius r and angle theta (counterclockwise from the x axis): x + iy = re^(i*theta) r = sqrt(x^2 + y^2) theta = arctan(y/x) Then, using Euler's formula, (x+iy)^k = r^k e^(i*k*theta) = r^k(cos(k*theta) + i*sin(k*theta)) Here are some related pages from our archives: Square Roots in Complex Numbers http://mathforum.org/dr.math/problems/jurd11.6.97.html Proof of e^(iz) = cos(x) + isin(x) http://mathforum.org/dr.math/problems/graf.4.7.97.html Roots of Unity http://mathforum.org/dr.math/problems/volante4.18.97.html - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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