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

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<pre>
When we solve this quadratic equation:

{{{z^2+zi+(1-3i)=0}}}

by the quadratic formula and simplify, we get this:

{{{z = (-i +- sqrt(-5+12i) )/2 }}} 

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

{{{x+yi =sqrt(-5+12i)}}}

Squaring both sides:

{{{(x+yi)^2 = -5 + 12i}}}

{{{x^2+2xyi+y^2i^2 = -5 + 12i}}}

{{{x^2+2xyi+y^2(-1) = -5 + 12i}}}

{{{x^2+2xyi-y^2 = -5 + 12i}}}

Set real parts equal:

{{{x^2-y^2 = -5}}}

Set imaginary parts equal:

{{{2xy = 12}}}

{{{xy = 6}}}

Solve system:

{{{system(x^2-y^2 = -5,xy = 6)}}}

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(-5+12i)=x + yi}}} which can either be 2+3i or -2-3i.

Thus {{{"" +- sqrt(-5+12i)="" +- (2+3i)}}}

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So the solution to

{{{z^2+zi+(1-3i)=0}}}

which was:

{{{z = (-i +- sqrt(-5+12i) )/2 }}}

becomes

{{{z = (-i +- (2+3i) )/2 }}}

Using the +, that simplifies to 1+i

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

So those are the two solutions to the quadratic.

Edwin</pre>