Find all square roots of the number 3 + 4i.

To start we need to set up the equation described by the question. Our answers should be in the form a + bi (where a and b are real), and since any answers are the root of 3 + 4i we can write down the following equation. (a + bi)^2 = 3+4i. If we expand the bracket we get this, a^2 + 2abi + b^2*i^2 and since i^2 is -1 we have a^2 - b^2 + 2abi = 3 + 4i. Since the imaginary parts of both sides of the equation must be the same and the real parts must also be the same we have two separate equations. a^2 - b^2 = 3 and 2ab = 4. We need to solve these equations simultaneously to give a and b. By rearranging the second we can get b in terms of a. 2ab = 4 so ab =2 so b = 2/a. Then we can plug this into the first equation to give a^2 - 4/a^2 = 3. We can now multiply both sides by a^2 to get rid of the fraction a^4 - 4 = 3a^2  therefor a^4 - 3a^2 -4 = 0. We can factorise this equation by treating it as a quadratic with a^2 as the variable, this gives us(a^2 - 4)(a^2 + 1) = 0 thus a^2 = 4 or a^2 = -1 this means a = 2 or a = -2 or a = i or a = -i. Since we defined a and b as being real at the beginning we know a = 2 or a = -2. By using the equation 2ab = 4 we get b = 1 or -1. This gives us the roots 2 + i and -2 - i.