Question

Find w=a + bi = √z , where a and b are real numbers.
√-60i-11

Answer 1

you could use the same steps as in your other post, where I used De Moivre's Theorem
or, try this approach

let a+bi = √(-11 - 60i)
then (a+bi)^2 = -11 - 60i
a^2 + 2abi + b^2i^2 = -11 - 60i
a^2 - b^2 + 2abi = -11 - 60i
equate the real parts and equate the complex part
a^2 - b^2 = -11 and 2ab = -60
from 2ab = -60
ab = -30 ---> b = -30/a
sub into other equation....
a^2 - (-30/a)^2 = -11
a^4 - 900 + 11a^2 = 0
(a^2 - 25)(a^2 + 36) = 0
a = ±5 or a is imaginary, but in a + bi, the a and b are real

if a = 5, then b = -30/5 = -6
if a = -5, then b = 6

√(-11 - 60i) = 5 - 6i or -5 + 6i