Lesson: The Complex Exponential

Question 2

1 point


Repeated uses of the rule for multiplying complex numbers in polar form yields the following theorem, known as de Moivre’s Formula. Let z have polar form r(cosine(theta) + i(sine(theta))), with r not equal to 0. Then for any integer n, we have that z-to-the-n = (r-to-the-n)(cosine(n(theta)) + i(sine(n(theta)))).

To prove this, we will first prove the result for all non-negative integers n by induction. As a base case, note that if n = 0, then z-to-the-0 will equal 1, and we also have that (r-to-the-0)(cosine(0(theta)) + i(sine(0(theta)))) will equal 1(cosine(0) + i(sine(0))), which equals 1(1 + 0i), which equals 1. So we do, in fact, have that z-to-the-0 = (r-to-the-0)(cosine(0(theta)) + i(sine(0(theta)))).

Now, for our induction step, we assume that z-to-the-k = (r-to-the-k)(cosine(k(theta)) + i(sine(k(theta)))) for all complex numbers z. Then we have that z-to-the-(k+1) = z(z-to-the-k), which will equal (r(cosine(theta) + i(sine(theta))))((r-to-the-k)(cosine(k(theta)) + i(sine(k(theta))))). And using our formula for multiplication, we know that this equals r(r-to-the-k)(cosine(theta + k(theta)) + i(sine(theta + k(theta)))), and this equals (r-to-the-(k+1))(cosine((k+1)theta) + i(sine((k+1)theta))). So we have shown that z-to-the-n = (r-to-the-n)(cosine(n(theta)) + i(sine(n(theta)))) for any non-negative integer n.

But what if n is a negative integer? Then m = -n is a positive integer, and so we have that z-to-the-n = z-to-the-(-m), which equals (z-to-the-m)-to-the-(-1), which equals((r-to-the-m)(cosine(m(theta)) + i(sine(m(theta)))))-to-the-(-1). But this equals (1/(r-to-the-m))(cosine(-m(theta)) + i(sine(-m(theta)))). But of course, 1/(r-to-the-m) is r-to-the-n, and –m is n, so we still have a (cosine(n(theta)) + i(sine(n(theta)))). And so now we have that z-to-the-n = (r-to-the-n)(cosine(n(theta)) + i(sine(n(theta)))) for all integers n.

This, by the way, becomes the perfect time to go back and prove property 6 of the complex conjugate—that the conjugate of (z-to-the-n) = (the conjugate of z)-to-the-n. To prove this, we want to make use of both de Moivre’s Formula and Theorem 9.1.a, which says that if z = r(cosine(theta) + i(sine(theta))), then the conjugate of z equals r(cosine(-theta) + i(sine(-theta))). Well, then we’ll see that the conjugate of (z-to-the-n) equals the conjugate of ((r-to-the-n)(cosine(n(theta)) + i(sine(n(theta))))), which we now know is (r-to-the-n)(cosine(-n(theta)) + i(sine(-n(theta)))), which we now know is (r(cosine(-theta) + i(sine(-theta))))-to-the-n, which is (the conjugate of z)-to-the-n.

Before we pause to look at some examples, I want to push forward a bit more, as we will end up with an easier notation for polar form. Well, why didn’t we use the easier notation before? It’s because it comes from the fact that taking the nth power of a complex number results in multiplying the argument by n, not taking its power. There is another place where we’ve seen this kind of behaviour, and that’s when looking at exponentials. After all, in the real numbers, we would get that (x-to-the-a)-to-the-b = x-to-the-(ab), not x-to-the-(a-to-the-b). So, the function cosine(theta) + i(sine(theta)) behaves like an exponential function. We use this idea to define the following.

Euler’s Formula says that e-to-the-(i(theta)) = cosine(theta) + i(sine(theta)). And so for any complex number z = x + iy, we define e-to-the-z, which is equal to e-to-the-(x+iy), which equals (e-to-the-x)(e-to-the-(iy)).

So we can use Euler’s Formula to find a new way of writing a polar form for a complex number, since if z = r(cosine(theta) + i(sine(theta))), then we will have that z = r(e-to-the-(i(theta))). In this form, de Moivre’s Formula becomes that z-to-the-n = (r-to-the-n)(e-to-the-(in(theta))).

So let’s look at some examples of this. In the previous lecture, we found that (the square root of 2)(cosine(pi/4) + i(sine(pi/4))) is a polar form for 1 + i. We can now write this as (square root of 2)(e-to-the-(i(pi)/4)), and use it to calculate that (1 + i)-to-the-fourth will equal ((square root of 2)-to-the-fourth)(e-to-the-((4)(i(pi)/4))), which equals 4(e-to-the-(i(pi))). If we want, we can even put this back into standard form by computing that 4(e-to-the-(i(pi))) = 4(cosine(pi) + i(sine(pi))), which equals 4(-1 + 0i), which is -4.

Also in the previous lecture, we found that -3 – (root 3)i = (2(root 3))(cosine(2(pi)/3) + i(sine(2(pi)/3))), so now we can write that -3 – (root 3)i = (2(root 3))(e-to-the-(i(2(pi))/3)). And now we can compute that (-3 – (root 3))-to-the-fifth = ((2(root 3))-to-the-fifth)(e-to-the-((5)(i(2(pi))/3))). Well, this equals ((12-to-the-(1/2))-to-the-fifth)(e-to-the-(i(10(pi)/3))). We can write 12-to-the-(5/2) as 288(root 3), and we can subtract 2(pi) from 10(pi)/3 to get the equivalent argument of 4(pi)/3, and this gives us a final answer that (-3 – (root 3)i)-to-the-fifth = (288(root 3))(e-to-the-(i(4(pi)/3))). And if we want, we can put our answer back in standard form by computing that (288(root 3))(e-to-the-(i(4(pi)/3))) = (288(root 3))(cosine(4(pi)/3) + i(sine(4(pi)/3))), which equals (288(root 3))(-1/2 + i(-(root 3)/2)), and this equals -144(root 3) – 432i.

As one last example, let’s find (1 – 2i)-to-the-tenth. In the last lecture, we found that 1 – 2i is approximately (the square root of 5)(cosine(-1.11) + i(sine(-1.11))). So now we could say that 1 – 2i is approximately (square root of 5)(e-to-the-(-1.11i)). And now we’ll get that (1 – 2i)-to-the-tenth = ((square root of 5)-to-the-tenth)(e-to-the-((10)(-1.11i))). Well, this equals 3125(e-to-the-(-11.1i)). If we want, we can add multiples of 2(pi), which is approximately 6.28, to the -11.1 to put it into the usual –pi to 2(pi) range. Adding two multiples of 2(pi) gives us that (1 – 2i)-to-the-tenth is approximately 3125(e-to-the-(1.46i)). In this case, I would not recommend putting the answer back into standard form, as the rounding off done during our calculations significantly changes the answer, which is 237 + 3116i, by the way. If you are savvy with a calculator, you can do the entire calculation, from finding theta = (sine-inverse)(-2(root 5)) to computing 3125(cosine(10(theta))) and 3125(sine(10(theta))), on your calculator without using any approximation, and you will get this result.

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