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Finding the Roots of a Complex Number - Problem 2 2,356 views
You know that i is the square root of -1, but what are the square roots of i? We are going to find both of the square roots of i in this example. To do that we have to set up an equation.
The square roots of i are solutions to the equation z² equals i, where z is some complex number. We'll say z is r cosine theta plus i sine theta. Now, when we square that, we get r² cosine 2 theta plus i sine 2 theta, by Demoivre’s theorem. And that’s going to equals i, which I should also write in trig form. Well i has a modulus of 1, and it’s got an argument of pi over 2. That’s one argument that works.
You can also add 2 pi, or 4 pi, or 6 pi, or any even multiple of pi to pi over 2, and you will still get a workable argument. So I’m going to say cosine, actually let me write 1 times cosine pi over 2 plus i sine pi over 2. Now for these two sides to be equal, you need r² to equal 1. And you need 2 theta to equal pi over 2 plus 2n pi. So you could add any even multiple of pi. And that means theta is pi over 4 plus n pi. Now for r, normally when you are solving this equation, you write r equals plus or minus 1. Remember that the modulus of an imaginary number are complex number has to be positive, so we need r to equal 1.
So let’s take a look at the square roots, first for n equals 0. When n equals 0, theta, the arguments pi over 4. And so the square root is 1, 1 is the modulus, times cosine of pi over 4, plus i sine pi over 4. Sine and cosine of pi over 4, are root 2 over 2. So this just becomes root 2 over 2 plus i root 2 over 2. That’s one of the square roots of i.
Let’s find the other one. Because remember there will only be two. The square root right that’s the second root of i, so there will only be 2 distinct square roots.
When n equals 1, I get pi over 4 plus pi, that’s 4 pi over 4. So I get a total of 5 pi over 4 and that’s my argument. Z equals 1 times cosine of 5 pi over 4 plus i sine of 5 pi over 4. Cosine of 5 pi over 4 is negative root 2 over 2, and so is the sine of 5 pi over 4.
So I get negative root 2 over 2, minus i root 2 over 2, that’s the other square root of i. Square either of these two numbers and you’ll get this guy. Let’s take a look at the graph of the two square roots.
Remember that i has a modulus of 1, and therefore so does both of its square roots, modulus 1. So they are all the same distance away from the origin. Notice the argument of i, pi over 2, this guy is half the argument pi over 4. Notice also, the 2 square roots are 180 degrees apart. We’ve got 2 fold rotational symmetry. We always get this symmetry when we are looking at the roots of a complex number.
The roots, the square roots of i.