SOLUTION: Heya, I'd be really greatful if you could solve this question from calculus, complex numbers for me. Find the complex number u = x + yi, where x and y are real numbers, so that

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson  -> Lesson -> SOLUTION: Heya, I'd be really greatful if you could solve this question from calculus, complex numbers for me. Find the complex number u = x + yi, where x and y are real numbers, so that      Log On


   



Question 1083004: Heya, I'd be really greatful if you could solve this question
from calculus, complex numbers for me.
Find the complex number u = x + yi, where x and y are real
numbers, so that u^2 = -5 + 12i. Then solve the equation
z^2+zi+(1-3i)=0

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Heya, I'd be really greatful if you could solve this question
from calculus, complex numbers for me.
Find the complex number u = x + yi, where x and y are real
numbers, so that u^2 = -5 + 12i. Then solve the equation
z^2+zi+(1-3i)=0
-------------------------------------------
When we solve this quadratic equation:

z%5E2%2Bzi%2B%281-3i%29=0

by the quadratic formula and simplify, we get this:

z+=+%28-i+%2B-+sqrt%28-5%2B12i%29+%29%2F2+ 

To simplify further we need to find a simpler form for
the square root, say u = x + yi such that

x%2Byi+=sqrt%28-5%2B12i%29

Squaring both sides:

%28x%2Byi%29%5E2+=+-5+%2B+12i

x%5E2%2B2xyi%2By%5E2i%5E2+=+-5+%2B+12i

x%5E2%2B2xyi%2By%5E2%28-1%29+=+-5+%2B+12i

x%5E2%2B2xyi-y%5E2+=+-5+%2B+12i

Set real parts equal:

x%5E2-y%5E2+=+-5

Set imaginary parts equal:

2xy+=+12

xy+=+6

Solve system:

system%28x%5E2-y%5E2+=+-5%2Cxy+=+6%29

Solve the second for a letter , substitute it in the first,

and get two solutions (x,y) = (2,3) and (x,y) = (-2,-3).

So +sqrt%28-5%2B12i%29=x+%2B+yi which can either be 2+3i or -2-3i.

Thus %22%22+%2B-+sqrt%28-5%2B12i%29=%22%22+%2B-+%282%2B3i%29

--------------------------------------------------------

So the solution to

z%5E2%2Bzi%2B%281-3i%29=0

which was:

z+=+%28-i+%2B-+sqrt%28-5%2B12i%29+%29%2F2+

becomes

z+=+%28-i+%2B-+%282%2B3i%29+%29%2F2+

Using the +, that simplifies to 1+i

Using the -, that simplifies to -1-2i

So those are the two solutions to the quadratic.

Edwin