1. The problem statement, all variables and given/known data a) Find the modulus and argument of 6^(1/2) + 2^(1/2)i b) Solve the equation z^(3/4) = 6^(1/2) + 2^(1/2)i 2. Relevant equations 3. The attempt at a solution For part a) i used Pythagoras to find the modulus. ( (6^(1/2))^2 + (2^(1/2))^2 )^(1/2) = (6 + 2)^(1/2) = 8^(1/2) and to find the principle argument of z i said arctan(2^(1/2)/6^(1/2) = pi/6 I assume this is correct. for part b) z^(3/4) = 6^(1/2) + 2^(1/2)i Then z^(3/4) = 8^(1/2)(cos(pi/6 + 2kpi) +isin(pi/6 + 2kpi)) Then raising the R.H.S of the Equation to 4/3 to get z and Then using de Moivre's theorem. therefore z = 4(cos(4pi/18 + 8kpi/3) +isin(4pi/18 + 8kpi/3)) This is where i am stuck. As k [itex]\in Z [/itex] i can get a value of theta which is contained within -pi < theta <= pi when k = 0. if i assign a value of 1 to K then i get a value of [itex]\vartheta[/itex] which is greater than pi. the answer i get when K = 0 is wrong and there are 3 solutions. So i must be doing something wrong elsewhere. I'm sorry in advance if i have made a silly error.