The complex number -2sqrt(2) + 2sqrt(6)I can be expressed in the form r*exp(iTheta), where r>0 and -pi < theta <= pi. By using the form r*exp(iTheta) solve the equation z^5 = -2sqrt(2) + 2sqrt(6)i.

Using an argand diagram we can see that this (in fact all) complex numbers create a right angled triangle and using this fact we can find the modulus of this number. The modulus of this number is the 'r' in question and there is a very sleek proof of this theorem but this question doesn't require that so we will just use this fact and calculate our modulus for this number i.e. r = sqrt(32). Now with some simple manipulation (I will show a trick for this to be converted into sqrt(2)^n form during the interview), we know that 2^5 = 32 therefore sqrt(32) = 2^(5/2) = sqrt(2)^5. Next, we need to find the angle theta. To do that we just use simple trigonometry on the Argand diagram of this number (will show this in during interview) and find that tan(Theta) = -sqrt(3) (NOTE: be very careful of the signs, use Argand diagram for better visual understanding).Using the Trigonometric Table (very easy to create for rough work, can show how during the interview) or calculator we see that the angle Theta is one of two possibilities but looking at argand diagram we will see that it can only take one value which is Theta = 120 degrees = 2pi/3 radians. This implies that -2sqrt(2) + 2sqrt(6)i = sqrt(2)^5exp(2pi/3 i). Thus if we fix z^5 = sqrt(2)^5exp(2pi/3 i) = sqrt(2)^5*exp(i(2pi/3 + 2kpi)) for k in Naturals i.e k takes values from {0,1,2,3,4,5,...} and taking 5th root of the equation we will get 5 different solutions. The end result is a simple arithmetic manipulation which can be done manually or using calculator. So we get z in the form z = sqrt(2)*exp(i(2pi/3 + 2kpi)/5).