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

z = 2 5 ( cos ⁡ 297 ° + i sin ⁡ 297 ° ) z=2\sqrt5\left(\cos297\degree+i\sin297\degree\right) z = 2 5 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> ( cos 297° + i sin 297° )

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

So in this problem, we're asked to express the complex number z is equal to 2 minus 4 I in polar form. Now recall that for complex numbers in polar form, they're going to take this format. They're going to be r times the cosine of theta, plus I times the sine of theta. So our mission is to figure out what this r and this theta are so we can plug it into this equation. Now r is going to be the square root of x squared plus y squared, and theta is going to be the inverse tangent of y over x. Now in these equations, x is the real part of our complex number, and y is the imaginary part. So let's see how we can solve this. I'm first gonna start by figuring out what r is. Then r is gonna be the square root of the real part squared, which would be 2 squared, plus the imaginary part squared, which we can see is negative 4 squared. Now 2 squared, that comes up to 4, and negative 4 squared is positive 16. 4 plus 16 is 20, and this square root is actually going to break down Because the square root of 20 is the same thing as the square root of 2 times 2 times 5. And what I can do is I can pull this pair of twos out because this pair of twos would be squared. So So we can also write this as the square root of 2 squared times the square root of 5, and then this will cancel with the square there, giving us 2 times the square root of 5. This is just how we reduce radicals, but this right here would be our solution for r. So that's what the r value is going to be. Now next we need to figure out theta. Now theta is going to be the inverse tangent of y over x. So we're gonna have the inverse tangent of our imaginary part which is negative 4 divided by the real part, which is positive 2. Now this can actually reduce. This would be the same thing as the inverse tangent of just negative 2, because negative 4 divided by 2 is negative 2. If you type the inverse tangent of negative 2 into a calculator, you should get approximately negative 63 degrees, and make sure you're in degree mode when you do this. Now the question we have to ask ourselves is, is this the angle we're looking for? Because recall that when looking for the total angle, you need to account for what quadrant you end up in. So what I'm going to do is draw a graph over here. And on this graph, we'll have the real axis and we'll have the imaginary axis. Now what I can see is that on the real axis, we're going to be 1, 2 units to the right and on the imaginary axis we're going to be negative 4. So 1, 2, 3, 4 units down. So our point should end up somewhere over here. Now of the 4 quadrants, notice how we ended up in quadrant 4. And for quadrant 4, this would not be the full angle solution. Because what we're really looking for, if we imagine this right here is R, we're trying to figure out what this total angle is gonna be if we start here on the positive x axis and loop all the way around. So if we end up in quadrant 4, we need to take whatever our angle is and add 360 degrees. So what I can do is go over here and add 360 degrees to this angle that I calculated. So we have negative 63 degrees plus 360 degrees, which comes out to 297 degrees. So this is actually the solution for our angle. This is the total angle that we get. So now I can plug everything in to our equation for polar form. So we're going to have that z is equal to r, which we figured out is 2 times the square root of 5, then that's going to be multiplied by the cosine of our angle, which is 297 degrees, plus I times the sine of our angle 297 degrees. So this right here is going to be the solution to the problem, and that is so you can solve these situations where we're dealing with polar form, and we also have an angle that doesn't end up somewhere in quadrant 1. So this is the solution. Hope you found this video helpful. Thanks for watching.