SOLUTION: How would you write {{{sqrt(3-4i)}}} in standard form

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Question 1083651: How would you write sqrt%283-4i%29 in standard form

Found 2 solutions by Edwin McCravy, rothauserc:
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Let sqrt%283-4i%29%22%22=%22%22x%2Byi, where x and y are real.

Since complex imaginary numbers are neither positive nor
negative, there is no rule about using the radical with
complex imaginary numbers as there is for real numbers.
So I suppose they want both square roots of 3-4i.

square both sides

3-4i%22%22=%22%22x%5E2-y%5E2%2B2xyi

Equate real parts:

x%5E2-y%5E2=3

Equate imaginary parts:

2xy=-4

Solve the system of equations:

system%28x%5E2-y%5E2=3%2C2xy=-4%29

Solve the second for y and substitute in the first,
and solve for the real soultions for x, then substitute
in one of the equations to get the real solutions for y.

x+=+%22%22+%2B-+2

y+=+%22%22+%2B-+1

At first there are four potential solutions:

2+i, 2-i, -2+i, -2-i.

However, to check for extraneous answers, we find that

by squaring each of them, we find that only 2-i and -2+i 
are solutions.

Edwin

Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
standard form of a complex number is a + bi, where a and b are real numbers and i is the square root(-1)
:
sqrt(3 - 4i) = a+bi
:
square both sides of =
:
3 - 4i = (a+bi)^2
3 - 4i = (a^2 + 2abi -b^2)
:
we have two equations in two unknowns
:
1) 3 = a^2 - b^2
2) -4i = 2abi
:
solve equation 2 for a
:
a = -4i / 2bi = -2/b
:
substitute for a in equation 1 and solve for b
:
b^4 +3b^2 -4 = 0
:
let u = b^2, then
:
u^2 +3u -4 = 0
:
factor polynomial
:
(u+4) * (u-1) = 0
:
u = -4 and u = 1
:
b^2 = -4 and b^2 = 1
:
b = 2i, -2i, 1, -1, b is real so reject 2i and -2i
:
a = -2/b
:
a = -2 or 2
:
********************************
standard form is 2 - i or -2 + i
********************************
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