Polar Form of Complex Numbers - Online Tutor, Practice Problems & Exam Prep

Complex numbers can be represented in polar form as \( r(\cos \theta + i \sin \theta) \), where \( r \) is the distance from the origin and \( \theta \) is the angle with the real axis. To find \( r \), use the equation \( r = \sqrt{x^2 + y^2} \), and for \( \theta \), apply \( \tan \theta = \frac{y}{x} \). Converting back to rectangular form involves distributing \( r \) into \( r(\cos \theta + i \sin \theta) \) to yield \( x + yi \). Understanding these conversions is essential for working with complex numbers effectively.

Complex Numbers In Polar Form

Hey there, everyone, and welcome back. So in the last video, we got introduced to how you could take these complex numbers in the form x+yi, and how you could plot them on these graphs. Now, what we're going to be talking about in this video is a new way that you can represent complex numbers using something called polar form. And polar form will take a complex number, and it's a way that you can write it with respect to the distance from the origin, which we call r, and the angle that we make with the real axis, which we call θ. Now, this might sound like something that's super scary and complicated, but don't sweat it. Because even though there are a few steps to solving these types of problems, once you know the equations, it's actually pretty intuitive. So let's just go ahead and get right into an example to see how we can deal with this polar form.

So let's say in this example, we're asked to write the complex number 4+3i in polar form, and we can see that we have the point right there. Now if I want to do this, this is the equation you need. x+yi, your complex number, is going to be equal to r×cos⁡(θ), and remember r is the distance from the origin, plus i times the sine of θ, where θ right here is the angle that we make with the real axis. So as long as you're able to figure out what r and these θ variables are, you can solve any type of polar form problem. But, of course, the question becomes what exactly are these variables? Well, let's talk about how we can calculate them.

So the way that you can do this is you can recognize that we actually have a familiar shape here that forms. Notice we have a right triangle in this situation, and we've dealt with right triangles many times in this course. Now if I want to calculate this long side of the triangle r, this is equivalent to the hypotenuse, and we know that for finding the hypotenuse we can just use the Pythagorean theorem. C is equal to the square root of a2+b2, where a and b are these two sides. So if I'm dealing with R as the hypotenuse, and we have y and x as these sides, I could also say that R is equal to x2+y2. Now likewise, if I want to find this angle, I can again use this right triangle as an example. Recall that the tangent is equal to opposite over adjacent. So going back to this triangle, I can see that the tangent of our angle θ is equal to the opposite side y divided by the adjacent side x. So these are the 2 equations that you need, which will allow you to find r and θ, and then once you find those, you can plug everything in to your polar form up here and solve the problem. So let's go ahead and do this for the example that we have.

Well, I'm first going to use this equation to find r. So we have that r is equal to the square root of x2+y2. I can see that x is going to be the real part, which is 4 squared, plus y, which is the imaginary part, 3 squared. Now the square root of 4 squared plus 3 squared, this turns out to be equal to 5. Now we need to find θ. Well, I can see that θ is going to be y over the tangent of θ is going to be y over x. So I can see that the tangent of θ is equal to y, the imaginary part 3, divided by x, the real part 4. Now what I can do is use the inverse tangent to find θ because θ is going to be the inverse tangent of our fraction here 3 over 4. That's what happens we take the inverse tangent on both sides of this equation. If you go ahead and plug this into a calculator, assuming you're in degree mode, you should get approximately 37 degrees as your answer. So that is the angle θ. So we now have our angle θ and we have r, so now I just need to put everything into this polar form equation. So we're going to have r which is 5, we figured that out right there, and that's going to be multiplied by the cosine of our angle 37 degrees, plus I times the sine of our angle, which is again, 37 degrees. So this right here would be the complex number in polar form. As you can see, we have this form up here, so we did this problem correctly.

So that is how you can solve these types of problems where you're trying to convert to this new polar form. Now, something else I want to mention about this is that this r that we calculated, this is always straightforward. You just use the Pythagorean theorem. But this θ is not always going to be very straightforward because the θ that you get, the angle, is going to depend on the quadrant you end up in. Now what do I mean by quadrant? Well, when it comes to your graph, there are 4 quadrants you can end up in. Now in this situation, we ended up in quadrant 1 because you can see that's the only quadrant we

Express the complex number $z=7+11i$ in polar form.

• A. $z=\sqrt{170}(\cos 58°+i\sin 58°)$
• B. $\sqrt{170}(\cos 58°+\sin 58°)$
• C. $z=\sqrt{170}+i58°$
• D. $z=170(\cos 58°+i\sin 58°)$

Express the complex number $z=2-4i$ in the polar form.

• A. $z=2\sqrt5\left(\sin297\degree+i\cos297\degree\right)$
• B. $z=2\sqrt5\left(\sin63\degree+i\cos63\degree\right)$
• C. $z=2\sqrt5\left(\cos297\degree+i\sin297\degree\right)$
• D. $z=2\sqrt5\left(\cos63\degree-i\sin63\degree\right)$

Express the complex number $z=1-\frac{\sqrt{3}}{3}i$ in polar form.

• A. $z=\frac{2\sqrt{3}}{3}(\cos \frac{\pi }{6}-i\sin \frac{\pi }{6})$
• B. $z=\frac{2\sqrt{3}}{3}(\cos \frac{13\pi }{6}+i\sin \frac{13\pi }{6})$
• C. $z=\frac{2}{\sqrt{3}}(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6})$
• D. $z=\frac{2\sqrt{3}}{3}(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6})$

Converting Complex Numbers from Polar to Rectangular Form

Welcome back, everyone. So in the previous video, we learned about how you could write complex numbers in polar form, which looks like this. Now, in this video, we're actually going to see how you can take complex numbers in this polar form and convert them back to their original form \(x + yi\). And this form, \(x + yi\), we actually call rectangular form. If this sounds complicated or abstract, don't sweat it. Because I think you're going to find that converting from polar to rectangular form is actually very straightforward. It's even more straightforward going in this direction than converting to polar form in the first place, in my opinion. And I think you might find that too. So without further ado, let's just go ahead and jump right into things.

Now let's say you have this complex number, and it's clear that we have polar form here since we see that we have an \(r\) value and we have an angle. If we want to take this and convert it back to rectangular form or just convert it there in the first place, all you really need to do is take whatever your \(r\) value is, and you need to distribute it. And that's really the main strategy for converting back to rectangular form. You distribute the \(r\)'s and then you evaluate whatever numbers you have, and that will tell you what your rectangular form is. So let's go ahead and try it right here.

We're given the example where we're asked to convert the complex number from forward to rectangular form and to identify the \(x\) and \(y\) values, or basically the real part and the imaginary part. So what I'm going to do is take this 5 and just distribute it into these parentheses. So if I do this, we're going to have \(5 \times \cos(37\degree)\) plus \(i \times 5 \times \sin(37\degree)\), and there's our distribution step right there. Now, \(5 \times \cos(37\degree)\) you can actually put that into a calculator. And if you do this, this should approximately come out to be 4. Now if you take \(5 \times \sin(37\degree)\) and you put that into a calculator you should get about 3. So \(4 + 3i\) is the complex number we get. And as we can see, the real part is 4 and the imaginary part is 3. So this here is a complex number and these are the solutions for \(x\) and \(y\). So as you can see, it's a very straightforward, very quick process to do this conversion.

Now to make sure that we know how to do this well, let's try one more example where this time we're dealing with radians. So as to convert the complex number from polar to rectangular form and as you can see here, we have our angle in radians this time. But, thankfully, we can use the same strategy where we just distribute the \(r\) value. So I'm first going to start by taking this 8, and I'm going to distribute it into these parentheses. So we're going to have \(8 \times \cos\left(\frac{\pi}{6}\right)\) minus \(i \times 8 \times \sin\left(\frac{\pi}{6}\right)\). So that's our first step. Now, from here, you have a couple of options. You could take the \(8 \times \cos\left(\frac{\pi}{6}\right)\) and \(8 \times \sin\left(\frac{\pi}{6}\right)\), and you could plug that into your calculator, but there's something we should recognize here. Both the cosine and the sine of \(\frac{\pi}{6}\) are actually values we can find on the unit circle. For the cosine, the cosine of \(\frac{\pi}{6}\) is the square root of 3 over 2. And the sine of \(\frac{\pi}{6}\) is actually 1 half. So if I rewrite this complex number which I'll do over here, we're going to have that \(z\) is equal to \(8\) multiplied by this value, \(\sqrt{3}/2\), And then this is going to be minus, and then we're going to have \(i\), and then that's going to be times \(8\), that's going to be multiplied by this quantity which is \(1/2\). So all I need to do is evaluate the numbers that I see. As you can see, I didn't even need to use my calculator for this one. So we're going to have that \(z\) is equal to \(8 \times \sqrt{3}/2\). Now, \(8\) divided by \(2\) is \(4\), so we're going to have \(4\sqrt{3}\), that's gonna be minus, then we'll have \(i\), then \(8\) divided by \(2\) is \(4\), so that's gonna be \(i \times 4\). So what our complex number becomes is \(4\sqrt{3} - 4i\). And this right here is our solution for \(z\) and that is the complex number in rectangular form. So as you can see, it's a very straightforward process. Just distribute and figure out what the sines and cosines are. You can do this on a calculator or with a unit circle, and then you can get your final answer. So that is how you can solve these types of problems. Hope you found this video helpful. Thanks for watching.

Convert the complex number $z=12\left(\cos90\degree+i・\sin90\degree\right)$ from polar to rectangular form.

• A. $z=12i$
• B. $z=12$
• C. $z=1$
• D. $z=i$

Convert the complex number $z=\sqrt2\left(\cos\frac{7\pi}{4}+i・\sin\frac{7\pi}{4}\right)$ from polar to rectangular form.

• A. $z=\sqrt2-i\sqrt2$
• B. $z=1-i$
• C. $z=2-2i$
• D. $z=-i$