 In this video, we provide the solution to question number 17 from the practice final exam for math 1060, in which case we're asked to find all the square roots of the complex number 1 minus i times the square root of 3. Now our best bet to compute these square roots is to put this complex number into polar form first. So let's say this complex number is z, then what would be the modulus of z? The modulus would look like the square root of 1 squared plus the square root of 3 squared. So we end up with the square root of 1 plus 3, which is the square root of 4. And so the square root of 4 turns out to be a 2. And so we get the modulus here is going to be 2. The next thing to do is to compute the argument here. So notice that the real part is positive, the imaginary part is negative. So our complex number is going to terminate in the fourth quadrant here. And so then thinking about like a tangent ratio, tangent theta is equal to, we're going to negative root 3 over 1 right here. So negative root 3, this occurs when you have a negative root 3 over 2 over 1 half. So that's going to happen. So when does cosine equal 1 half and sine equals negative root 3 over 2? Of course this is going to be in the fourth quadrant here. Well sine is equal to root 3 over 2 at 60 degrees in the first quadrant. So we're looking for the angle that references 60 degrees in the fourth quadrant. So we're looking for 300 degrees or in other words we're trying to get theta equals 5 pi over 3. Now when it comes to the arguments here when you're taking complex roots, we have to realize that anything coterminal to that would also be acceptable. There is a change of period that happens here and so this is important to recognize. So z is going to equal if we put into its polar form 2 times e to the i times 5 pi thirds plus 2 pi k. So this is our complex number in polar form. That's the hardest part of the problem. The next step is then to take square root. If we take z to the 1 half power that's what we're going to do here. We're going to get the square root of 2 and then we're going to take e to the i and we're going to divide this by 2. So we get 5 pi thirds plus 2 pi k and we divide this angle by 2. So root 2 times e to the i we're going to get 5 pi over 6 plus pi k and so there's going to be two options here. There's the option where you take k to be 0 and k to be 1. So the square roots of z you're going to end up with the square root of 2 times e to the i 5 pi sixth and then you have root 2 times e to the i. We're going to add pi to that so we end up with 11 pi sixth like so. All right and so now we need to compute these complex numbers using this angle. So we have the square root of 2 times cosine of 5 pi sixth plus i sine of 5 pi sixth that's the first square root. The second square root will be similar to the same modulus square root of 2 but we're going to cosine this case of 11 pi sixth and i excuse me i sine of 11 pi sixth like so. So notice 5 pi sixth and 11 pi sixth both reference pi sixth or 30 degrees in the first quadrant. Now 5 pi sixth is in the second quadrant so cosine would be negative in that situation cosine of pi sixth. Sine would be positive. For the second one of course 11 pi sixth is in the fourth quadrant so cosine would be positive in that situation for which then you get pi sixth and then the next one you're actually getting negative. Sine is negative in the fourth quadrant so you get negative i sine of pi sixth and that's not too surprising. Whenever you take square roots whatever the principal square root is the other square root will be its additive inverse. So notice how the sines are different negative negative versus positive when you compare the two. So now we have to do cosine of pi sixth which would be root 3 over 2. Sine of pi sixth is one half and so doing as such we get the square root of 2 times negative root 3 over 2 plus i over 2 that's the first square root and then the second square will be the square root of 2 times well same basic numbers the sines are different though 3 over 2 minus i over 2 like so. So we potentially could stop there but let's distribute the square root of 2 through on this and so for the first one we get negative square root of 6 over 2 plus we're going to get i root 2 over 2 that's the first square root of z and then the second one's gonna be this again just additive inverse the square root of 6 over 2 minus i root 2 over 2 and so this then gives us the two complex square roots of the number 1 minus i square root of 3.