SOLUTION: Solve the equation z^2 + (2i - 3)z + 5 - i = 0.

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Question 1123447: Solve the equation z^2 + (2i - 3)z + 5 - i = 0.
Found 2 solutions by greenestamps, solver91311:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


In the quadratic equation

ax^2+bx+c = 0

the sum of the roots is -b/a and the product is c/a.

This holds for quadratic equations with complex coefficients.

With a=1 in this equation, the sum of the roots is -(2i-3) = 3-2i and the product is 5-i.

Let the roots be p+qi and r+si. Then their product is

(pr-qs)+(ps+qr)i

and their sum is

(p+r)+(q+s)i

So we have four equations in p, q, r, and s:

pr-qs = 5; ps+qr = -1
and
p+r= 3; q+s = -2

You can solve the problem by playing around with small integers, trying to find solutions for the second pair of equations that give correct results in the first pair of equations.

I leave that to you as an exercise.

Another way to solve the problem is by using the quadratic formula -- which also holds for quadratic equations with complex coefficients.

Indeed, many students who are not comfortable with factoring techniques always go straight to the quadratic formula for solving any quadratic equation.

For this equation, the quadratic formula gives us the two roots as

%28%283-2i%29+%2B+sqrt%28%282i-3%29%5E2-4%281%29%285-1%29%29%29%2F2 and %28%283-2i%29+-+sqrt%28%282i-3%29%5E2-4%281%29%285-1%29%29%29%2F2
%28%283-2i%29+%2B+sqrt%28%285-12i%29-%2820-4i%29%29%2F2%29 and %28%283-2i%29+-+sqrt%28%285-12i%29-%2820-4i%29%29%29%2F2
%28%283-2i%29+%2B+sqrt%28-15-8i%29%29%2F2 and %28%283-2i%29+-+sqrt%28-15-8i%29%29%2F2
%28%283-2i%29+%2B+%281-4i%29%29%2F2 and %28%283-2i%29+-+%281-4i%29%29%2F2
%282-3i%29 and %281%2Bi%29

Of course those are the two roots you should end up with if you try the trial-and-error method suggested above.

So...

Answer: The two roots are (2-3i) and (1+i)

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
















As an interim step, we need



















Either way:



Using







You get the same result if you use . Verification left as an exercise for the student.


John

My calculator said it, I believe it, that settles it