Hello from MrBillDoesMath!
Answer:
4 + i
Discussion:
Let's verify the square root given above:
(4 + i) ^ 2= 4^2 + i^2 + 8i = 16 - 1 + 81 = 15 +8i
so (4+i) is indeed a square root.
How did I find this? Suppose that
sqrt(15 + 8i) = a + bi
Squaring both sides gives
15 + 8i = a^2 + 2abi + b^2*i^2 or
15 + 8i = a^2 + 2abi + b^2 (-1) or
15 + 8i = (a^2 - b^2) + (2ab)i.
Equating real and imaginary parts gives:
a^2 - b^2 = 15 and 2ab = 8
Now 2ab= 8 => b = 8/(2a) = 4/a. Substitute this value in the equation
a^2 - b^2 = 15 to get
a^2 - (4/a)^2 = 15. Multiply both sides by a^2
(a^2) * (a^2) - (4/a)^2 * (a^2) = 15 * a^2 or
(a^4) - 16 = 15 a^2 or rearranging
(a^2) ^2 - 15 (a^2) - 16 = 0.
This is a quadratic equation in a^2! Use the quadratic formula to solve for
a^2 . You will get that a^2 = -1 ( no way! ) or a^2 = 16. If a^2 = 16, then
a = 4 and from 2ab = 8, 2(4)b = 8, or 8b = 8 or b = 1.
I am now officially tired...... hope this helps.
Thank you,
MrB
