What are Polar Coordinates?

 You probably already know what a coordinate is. Normally you would see a coordinate in the form of (x,y). For example the point (2,2) would look like this on a cartesian plane.

On a complex plane, you would write it in the form a + bi, where a is the real number and b is imaginary. For example the point 2 + 2i would look like this on a complex plane.

However, there is another way to write a coordinate on a complex plane. This is called a polar coordinate and it appears in the form of r(cos(x) + isin(x)). (Note: Normally x would be represented with theta, but you can’t type the symbol for theta on this blog, so I’m replacing it with x. The images with the graphs will have the symbol written as theta.)

Let’s imagine the complex plane again with the point 2 + 2i. If I tell you the exact number of degrees you need to rotate from the origin (x) and the distance from the origin to the coordinate point (r), you can find where the coordinate point is on the complex plane.

Let’s try a different problem. If I tell you that a point is rotated thirty degrees from the origin and is 3 units away, what would the position look like on a complex plane?

Or if I rotate by 60 degrees and I’m 4 units away from the origin.

If you put this in polar form, it would be 4(cos(60) + isin(60)). The real question is why would polar form be this? Why not just state x = 60 and r = 4?

The answer to that is, if you multiply it out, you will get the coordinate in the form a + bi.

4(cos(60) + isin(60)) = 4(cos(60)) + 4(isin(60))

Now you can multiply each of the components separately.

 

4 (cos(60)) = 2

4(isin(60)) ~ 3.46i

This gives you the result 2 + 3.46i. If we look at the graph again,

we can see that this is the point 2 + 3.46i.

So, why does this work? To figure this out let’s first look at what r in the polar coordinate is.

Let’s think of this as a triangle with r being the hypotenuse and the horizontal length and vertical length being a and b respectively.

Now let’s try to put r in terms of a and b. Since this is a right triangle, we can use the pythagorean theorem.

r2=a2+b2

This means:

r = sqrt(a2+b2)

Putting this back into polar form will give you:

sqrt(a2+b2)*(cos(x)+isin(x))

The next step is to put cos(x) and sin(x) in terms of a and b.

If we go back to the basics of trigonometry, we know that cos = adjacent / hypotenuse. This means cos(x) = a / r. We can do the same thing for sin. Sin = opposite / hypotenuse, which means sin(x) = b / r

Now let’s put this back into polar form. 

sqrt(a2+b2)*(a / r+i (b / r))

We know r =sqrt(a2+b2) , so we can rewrite that as:

sqrt(a2+b2)*(a /sqrt(a2+b2) +i (b / sqrt(a2+b2))

If you simplify this, you get:

sqrt(a2+b2)*(a /sqrt(a2+b2 )=a

sqrt(a2+b2)*i(b /sqrt(a2+b2 )=bi

So all together it’s equal to a + bi.