In order to convert to polar form, remember that
x + yi = r[cos(a) + isin(a)],
in which r, your magnitude, is sqrt[x^2 + y^2]
and the reference angle of your argument, a, is arctan(|y/x|) (i.e., tan inverse of the absolute value of y/x... using absolute value gives us a positive acute angle, which is our reference angle).
Then, the signs of x & y in the original determines the quadrant (and, thereby, the actual value) of your argument
r = sqrt[(-5)^2 + (5sqrt3)^2] = sqrt[25 + 75] = sqrt = 10
reference angle of a = arctan(|5sqrt3/-5|) = arctan(sqrt3)... Remember, arctan(sqrt3) is the angle whose tangent is sqrt3. Tan of 60degrees (or pi/3 radians) is sqrt3. So, reference angle is 60degrees or pi/3 radians.
Since, x is negative and y is positive, our quadrant is the second; our actual argument becomes 180 - 60 degrees (or pi - pi/3 radians)... This simplifies to 120 degrees or 2pi/3 radians.
So, polar form is 10(cos(120degrees) + isin(120degrees)) or
10(cos(2pi/3radians) + isin(2pi/3radians)).