Cube Square

How do I find the square root of 3 - 4i and the cube root of 1 - i?

Hi, plz help! I do not know exactly how and how / If I can express in terms of cosx + isinx? How can I find: 1. The square root of (3 - 4i) 2. The square root of (1 - i)

I will do for you LEAT first and second to implement what has been done in the first sqrt! (3 - 4i) = a + bi 3 - 4i = (a + b) ^ 2 3 - 4i = (a + ^ ^ 2-B 2abi 2) SO 3 = A ^ 2-b ^ 2-4i = 4i = 2abi self-2abi a = 2 / b, then if 3 = a ^ 2-b ^ 2 subst. a = 2 / b 3 = 4 / b ^ 2 - b ^ 2 solve for bb ^ 2 4 / b ^ 2 + 3 = 0 b ^ 4 + 3 b ^ 2-4 = 0 Now factor (b ^ 2 +4) (b ^ 2 -1) = 0 roots 2i,-2i, 1, -1 may not be used 2i-2i or b = -1 is now = 1 or 2 / B to a 2/-1 = = -2 or 2 = / 1 = 2 for the form a + bi answers might be (2 - 1i) or (-2 +1 i)