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

z = 170 ( cos ⁡ 58 ° + i sin ⁡ 58 ° ) z=\sqrt{170}\left(\cos58\degree+i\sin58\degree\right) z = 170 <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 58° + i sin 58° )

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

Let's give this problem a try. So here we're asked to express the complex number z is equal to 7+11i in polar form. Now recall that polar form looks something like this. Our complex number is going to be r times the cosine of theta, plus I times the sine of theta. That's polar form for complex numbers. Now r and theta are both things that we can calculate. R is going to be the square root of x squared plus y squared, where x and y are the real and imaginary part of our complex number respectively. Now theta is going to be the inverse tangent of y over x. So we can use both these equations to figure out what the r and theta value are which will allow us to find the equation, the complex number in polar form. So let's go ahead and do this. First, I'm gonna start by calculating r. R is going to be the square root of the real part squared plus the imaginary part squared. The real part I can see is 7 and the imaginary part is 11. Now 7 squared is 49, and then 11 squared is a 121. 49 +121 comes out to a 170. Now it turns out this radical doesn't break down any farther, so this right here is the answer for r. Now next, we need to calculate theta. Theta is going to be the inverse tangent of y over x. Now since y is the imaginary part that's gonna be 11, and x is going to be 7, the real part. Now the inverse tangent of 11 over 7, if you check that on a calculator, it comes out to approximately 58 degrees. So this would be our answer for theta but we need to actually confirm this. Because you may recall that this is not always going to be our total angle it depends on what quadrant your complex number ends up in. So So if I go ahead and draw a graph over here, we'll have the real part and the imaginary part. Now I can see that we have positive 7 for the real parts we're gonna be somewhere over here We have positive 11 for the imaginary part somewhere up here, meaning we're going to end up over here in quadrant 1. Now of the 4 quadrants, if we end up in quadrant 1, then this inverse tangent equation will calculate the angle that we need. So we actually have the angle, we have the r value, now all we need to do is plug everything in to polar form. So So we're going to have that z is equal to r which we calculated to be the square root of 170, then that's going to be multiplied by the cosine of our angle 58 degrees plus I times the sine of our angle, which is 58 degrees. So this right here is the solution for our complex number in polar form, and that's how you can solve these types of problems. So hope you found this video helpful. Thanks for watching.

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

z = 170 ( cos ⁡ 58 ° + i sin ⁡ 58 ° ) z=\sqrt{170}\left(\cos58\degree+i\sin58\degree\right) z = 170 <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 58° + i sin 58° )

170 ( cos ⁡ 58 ° + sin ⁡ 58 ° ) \sqrt{170}\left(\cos58\degree+\sin58\degree\right) 170 <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 58° + sin 58° )

z = 170 + i 58 ° z=\sqrt{170}+i58\degree z = 170 <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"> + i 58°