what is the square root of i, if i=x+yi? what is the square root of 1-i? i'm getting problems like these in whichĘI doĘnot understand. Thank You for your help, Arlene Hi Arlene, Every complex number z can be written z = x + iy where x and y are real numbers and i is the imaginary number that satisfies i 2 = -1. x is called the real part of z and y is the imiginary part of z. In particular i = x + iy where x = 0 and y = 1. Suppose that x and y are real numbers and x + iy is the square root of i, then (x + iy) 2 = i Squaring x + iy, remembering that i 2 = -1, gives x 2 - y 2 + 2ixy = i Since x 2 - y 2 + 2ixy and i are the same complex number their real and imaginary parts are equal. That is x 2 - y 2 = 0 and 2xy = 1 I want to solve these two equations for x and y. From the second equation y = 1/2x and thus, by substituting into the first equation, x 2 - (1/2x) 2 = 0 Solve this equation for x (you will get two solutions) and then use y = 1/2x to find the corresponding values of y. This will give you two complex numbers that are square roots of i. Finding the square root of 1 - i is similar. Suppose that (x + iy) 2 = 1 - i Squaring gives x 2 - y 2 + 2ixy = 1 - i Again set the real and imininary parts equal and solve for x and y I hope this helps, Harley Go to Math Central