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 which╩I do╩not 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
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