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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When we solve this quadratic equation:
by the quadratic formula and simplify, we get this:
To simplify further we need to find a simpler form for
the square root, say u = x + yi such that
Squaring both sides:
Set real parts equal:
Set imaginary parts equal:
Solve system:
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 which can either be 2+3i or -2-3i.
Thus
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So the solution to
which was:
becomes
Using the +, that simplifies to 1+i
Using the -, that simplifies to -1-2i
So those are the two solutions to the quadratic.
Edwin