# Lesson: Polar Form

1 point

## Transcript

In the previous lecture, we saw that we can visualize a complex number as a point in the complex plane. This turns out to be remarkably useful, but we need to think about things a bit differently. Instead of thinking of the point (x, y) in terms of its x1 and x2 components, we instead want to describe a point in terms of its distance from the origin and how much it is rotated away from the positive real axis.

Given a complex number z = x + yi, we define the modulus of z, denoted such as absolute value of z, to be the real number r equal to the square root of (x-squared + y-squared). If the modulus of z does not equal 0, let theta be the angle measure counterclockwise from the positive x-axis such that x would equal r(cosine(theta)) and y would equal r(sine(theta)). The angle theta is called an argument of z.

Note that we say an argument of z, since theta is only unique up to a multiple of 2(pi). As such, every complex number has an infinite number of arguments. We actually will make use of this fact. But for now, we’ll simply state that if r is the modulus of z, and if theta is an argument for z, then a polar form for z is r(cosine(theta) + i(sine(theta))).

So let’s look at some examples. To determine a polar form for 1 + i, we will first compute its modulus, r, equal to the square root of (1-squared + 1-squared), which equals the square root of 2. Next, we need to find a theta such that 1 = (the square root of 2)(cosine(theta)), and 1 = (the square root of 2)(sine(theta)). That is, we want cosine(theta) to equal 1/(square root 2), and sine(theta) to equal to 1/(square root of 2). Setting theta = pi/4 satisfies these equations, so a polar form for 1 + i is (the square root of 2)(cosine(pi/4) + i(sine(pi/4))).

The steps to determine a polar form for 1 – i are quite similar. First, we compute that the modulus r equals the square root of (1-squared + (-1)-squared)), which equals the square root of 2. Next, we need to find theta such that 1 = (the square root of 2)(cosine(theta)), and -1 = (the square root of 2)(sine(theta)). That is, we want cosine(theta) to equal to 1/(the square root of 2), and sine(theta) to equal -1/(the square root of 2). Now, since our cosine value is positive while our sine value is negative, we know we are looking for a theta in the fourth quadrant, so we can set theta = 7(pi)/4 to satisfy our equations. This gives the polar form (square root of 2)(cosine(7(pi)/4) + i(sine(7(pi)/4))) for 1 – i. But we might just as easily have chosen the theta value of –pi/4, though, giving us the polar form of (square root of 2)(cosine(-pi/4) + i(sine(-pi/4))) for the complex number 1 – i.

In general, you will do well to visualize the point x + iy when you are trying to determine an argument for it, instead of relying purely on calculations. This is particularly true when looking for a polar form for a purely real or purely imaginary number.

For example, let’s try to find the polar form for the numbers 4, -7, 2i, and -5i, first by graphing them on the complex plane. Now, just by looking at the graph, it becomes obvious that the polar form for 4 is 4(cosine(0) + i(sine(0))). A polar form for -7 is 7(cosine(pi) + i(sine(pi))). A polar form for 2i is 2(cosine(pi/2) + i(sine(pi/2))). And a polar form for -5i is 5(cosine(3(pi)/2) + i(sine(3(pi)/2))).

Theorem 9.1.a: If z = r(cosine(theta) + i(sine(theta))), then the conjugate of z equals r(cosine(-theta) + i(sine(-theta))). And since I’ve made it a theorem, I’ll go ahead and prove it for you. So we’ll let z = x + iy, so the conjugate of z is x – yi. Well, then the modulus for the conjugate of z will equal the square root of (x-squared + (-y)-quantity-squared), but of course, this is the same as the square root of (x-squared + y-squared), which is the modulus of z, which is r. So we see that the modulus of the conjugate of z is the same as the modulus for z.

Now, since z = r(cosine(theta)) + ir(sine(theta)), and using the fact that cosine(-theta) = cosine(theta), while sine(-theta) = -sine(theta), we’ll see that the conjugate of z will have to equal r(cosine(theta)) – ir(sine(theta)), which equals r(cosine(-theta)) + ir(sine(-theta)), which equals r(cosine(-theta) + i(sine(-theta))), as we desired.

We can use properties of cosine and sine to help us multiply and divide complex numbers as well. Let’s recall the following trigonometric identities, that the cosine(theta1 + theta2) will equal (cosine(theta1))(cosine(theta2)) – (sine(theta1))(sine(theta2)), and that sine(theta1 + theta2) = (sine(theta1))(cosine(theta2)) – (cosine(theta1))(sine(theta2)). So we can use them to get the following result, and I’ll note that this does appear in the textbook—it simply isn’t listed as a theorem there.

So I’ll call it Theorem 9.1.b. For any complex numbers z1 = r1(cosine(theta1) + i(sine(theta1))) and z2 = r2(cosine(theta2) + i(sine(theta2))), we have that z1z2 = r1r2(cosine(theta1 + theta2) + i(sine(theta1 + theta2))).

And again, since I’ve turned this into a theorem, I’ll go ahead and prove it for you. So again, we’ll let z1 and z2 be as described in the theorem, and then if we were to multiply z1z2, we can actually multiply their polar forms. First, we can go ahead and distribute our r1’s and our r2’s inside the individual polar forms, and now it simply becomes a question of distributing our multiplication. So we’ll end up with an r1r2(cosine(theta1))(cosine(theta2)) + i(r1r2)(cosine(theta1))(sine(theta2)) + i(r1r2)(sine(theta1))(cosine(theta2)) + (i-squared)(r1r2)(sine(theta1))(sine(theta2)). So let’s go ahead and pull all those (r1r2)’s out, and we can also start grouping things together. So, for example, we have our cosine(theta1) and our cosine(theta2), we could actually group with our + (i-squared)(r1r2)(sine(theta1))(sine(theta2)), but remembering that the i-squared becomes a -1, so this becomes the quantity (cosine(theta1))(cosine(theta2)) – (sine(theta1))(sine(theta2)). And then next, we’ll pull out our i terms, which leave us with the (cosine(theta1))(sine(theta2)) + (sine(theta1))(cosine(theta2)). But using our trig identities, we’ll see that this becomes (r1r2)(cosine(theta1 + theta2) + i(sine(theta1 + theta2))).

Next, we’ll do a theorem that is in the book. For any complex numbers z1, with polar form r1(cos(theta1) + i(sine(theta1))), and z2, with polar form r2(cos(theta2) + i(sine(theta2))), with z2 not equal to 0, we have that z1/z2 will equal (r1/r2)(cosine(theta1 – theta2) + i(sine(theta1 – theta2))).

Now, to prove this, let’s first remember that if z1/z2 = z3, then we have that z1 = z2z3. So we can see that z1/z2 will equal (r1/r2)(cosine(theta1 – theta2) + i(sine(theta1 – theta2))) by showing that z1 = z2 times this value. And we show this as follows. z2(r1/r2)(cosine(theta1 – theta2) + i(sine(theta1 – theta2))) will equal (r2(cosine(theta2) + i(sine(theta2)))) times our other value. But now, we’ll use the fact that we’re multiplying these two polar forms, and our Theorem 9.1.b, to see that this equals (r2(r1/r2))(cosine(theta2 + (theta1 – theta2)) + i(sine(theta2 + (theta1 – theta2)))), and of course, our r2’s cancel out, leaving us with an r1, and our theta2’s cancel out, leaving us with a cosine(theta1) and a sine(theta1), which is, of course, z1 as desired.

One special case of this theorem is the following. Let z = r(cosine(theta) + i(sine(theta))), with r not equal to 0. Then z-inverse will equal (1/r)(cosine(-theta) + i(sine(-theta))). This is simply using Theorem 9.1.3 with our z1 being the value 1 and our z2 being z.

Let’s look at some examples of using these theorems. So earlier, we found that a polar form for 1 + i is (the square root of 2)(cosine(pi/4) + i(sine(pi/4))). Now let’s find a polar form for –(root 3) + 3i. First, we can find that the modulus r equals the square root of ((-3)-quantity-squared + 3-squared), which is the square root of 12, or we could write that as 2(root 3). Next, we need to find theta such that the cosine(theta) = –(root 3)/(2(root 3)), which is -1/2, and sine(theta) = 3/(2(root 3)), which is (root 3)/2. Now, since our cosine value is negative and our sine value is positive, we know that theta is in the second quadrant, so we can use theta = 2(pi)/3 as an argument. So this means that (2(root 3))(cosine(2(pi)/3) + i(sine(2(pi)/3))) is the polar form for -3 + 3i.

At this point, we can use these polar forms to perform the following calculations. So (1 + i)(-3 + 3i) equals the product of their polar forms. But this is simply the product of the moduli, and then the sum of the arguments, so we get (2(root 6))(cosine(7(pi)/12) + i(sine(7(pi)/12))). Meanwhile, we can look at (1 + i)/(-(root 3) + 3i) by looking at the polar form for (1 + i) divided by the polar form for (-(root 3) + 3i). This means we’re going to divide the moduli and subtract the second argument from the first. So we’re left with (1/(root 6))(cosine(-pi/12) + i(sine(-pi/12))). What if, instead, we wanted to look at taking (-(root 3) + 3i) and dividing it by (1 + i)? Again, we’ll want to turn this into their polar forms and looking at that quotient, which means we’ll look at the quotient of the moduli, and we’ll subtract the second argument from the first. So we’re getting that it has a modulus of (root 6), and then times (cosine(pi/12) + i(sine(pi/12))). Now, note that this last result agrees with the result we would get if we considered (-(root 3) + 3i)/(1 + i) to be the inverse of (1 + i)/(-(root 3) + 3i).

Let’s use the polar form to calculate the product (1 – 2i)(-3 – 4i). Well, first we’ll need to find the polar form of (1 – 2i) and (-3 – 4i). Now, this is not as easy as in the previous example, as our arguments will not be one of our standard angles. Instead, we will use a calculator and decimal approximations for these calculations. So to find a polar form of (1 – 2i), we first note that the modulus r will equal the square root of (1-squared + (-2)-squared), which equals the square root of 5. And now we need to find theta such that the cosine(theta) = 1/(the square root of 5), and sine(theta) = -2/(the square root of 5). Now, since the cosine is positive and the sine is negative, we know that theta is in the fourth quadrant. Plugging theta = (sine-inverse)(-2/(root 5)) into a calculator will give us a value in the fourth quadrant, and so we will see that theta is approximately -1.11. And so we have found that a polar form of (1 – 2i) is (root 5)(cosine(-1.11) + i(sine(-1.11))).

To find a polar form of (-3 – 4i), we first note that the modulus for this will equal the square root of ((-3)-squared + (-4)-squared), which is the square root of (9 + 16), which equals 5. And now we need to find theta such that cosine(theta) = -3/5 and sine(theta) = -4/5. But since the cosine is negative and the sine is negative, we know that theta is in the third quadrant. Unfortunately, a calculator never outputs a value in the third quadrant. So we can either have that theta = –(cosine-inverse)(-3/5) or that theta = (cosine-inverse)(3/5) + pi. I’ll choose to do the first, which gives me that theta = -2.21. And so we have found that a polar form of (-3 – 4i) is 5(cosine(-2.21) + i(sine(-2.21))).

Now that we have polar forms for (1 – 2i) and (-3 – 4i), we can calculate that the product (1 – 2i)(-3 – 4i) will equal the product of their polar forms, which we now know equals the product of their moduli and the sum of their arguments, so we get (5(root 5))(cosine(-3.32) + i(sine(-3.32))). We can double-check our calculation by putting things back into standard form. With a healthy amount of round-off error, we do, in fact, see that (5(root 5))(cosine(-3.32) + i(sine(-3.32))) = -11 + 2i.

Now, given how much effort it is to find polar form, at this point, it probably does not seem worthwhile to put complex numbers into polar form simply to multiply and divide them. But in a couple of lectures, we will use polar form to find the nth roots of complex numbers, and this really is the best way to solve this type of question. So for now, simply consider this lecture a gentle introduction to polar form and calculations in polar form, but polar form is not going away.