Lesson Solution of the quadratic equation with complex coefficients on complex domain

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Solution of the quadratic equation with complex coefficients on complex domain


In this lesson I will explain you how to solve the quadratic equation
ax%5E2+%2B+bx+%2B+c+=+0,              (1)
with complex coefficients a, b and c on complex domain.
This means that the coefficients of the equation are complex numbers, and we look for roots among the set of complex numbers.

Reminder 1


Let me remind you that the solution of the quadratic equation with real coefficients on real domain was considered in the lesson
Introduction into Quadratic Equations of the module Quadratic Equation of this site, as well as in other lessons in that module.

The key formula for the solution of the quadratic equation is the quadratic formula
x+=+%28-b+%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29.      (2)
This formula was deduced in the lesson PROOF of quadratic formula by completing the square of the module Quadratic Equation of this site.
The method "completing the square" was used in this deduction.

The discriminant
d+=+b%5E2+-+4%2Aa%2Ac
plays the key role in solution of the quadratic equation.

If the discriminant value is positive then the equation has two real roots
x+=+%28-b+%2B-+sqrt%28d%29%29%2F%282a%29.                   (3)

If the discriminant value is equal to zero then the equation has only one real root
x+=+-b%2F%282a%29.                      (4)

If the discriminant value is negative then the equation has no real roots.

Reminder 2


The solution of the quadratic equation with real coefficients on complex domain was considered in the lesson
Solution of the quadratic equation with real coefficients on complex domain in the current module.

In that lesson it was shown that for real coefficients a, b and c the solutions of the quadratic equation (1) are real numbers
x+=+%28-b+%2B-+sqrt%28d%29%29%2F%282a%29                   (5)
if the discriminant d is non-negative, or complex (conjugate) numbers
x+=+%28-b%2F%282a%29%29+%2B-+red%28i%29%2A%28sqrt%28abs%28d%29%29%2F%282a%29%29      (6)
if the discriminant d is negative.

After these reminders the rest in this lesson is simple and straightforward.

First of all, the same method "completing the square" works for the quadratic equation (1) with complex coefficients also.
Therefore, the same quadratic formula is valid for the roots of the quadratic equations (1) with complex coefficients
x+=+%28-b+%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29      (7)
or, in the other form,
x+=+%28-b+%2B-+sqrt%28d%29%29%2F%282a%29                   (8)
where
d+=+b%5E2+-+4%2Aa%2Ac                 (9)
is the discriminant.

Now, when a, b and c are complex numbers, the discriminant d+=+b%5E2+-+4%2Aa%2Ac is the complex number also, and you can write it in the form
d+=+e+%2B+red%28i%29%2Af,
where e and f are real numbers.

It was shown in the lesson How to take a square root of a complex number of this module that
the square root of the complex number d+=+e+%2B+red%28i%29%2Af has two values.

The first value is the complex number
w%5B1%5D+=+r%5B1%5D+%2B+red%28i%29%2As%5B1%5D,
where
r%5B1%5D+=+sqrt%28%28e+%2B+sqrt%28e%5E2+%2B+f%5E2%29%29%2F2%29, s%5B1%5D = +/-sqrt%28%28sqrt%28e%5E2+%2B+f%5E2%29+-+e%29%2F2%29.

The second value is the complex number
w%5B2%5D+=+r%5B2%5D+%2B+red%28i%29%2As%5B2%5D,
where
r%5B2%5D+=+-sqrt%28%28e+%2B+sqrt%28e%5E2+%2B+f%5E2%29%29%2F2%29, s%5B2%5D = -/+sqrt%28%28sqrt%28e%5E2+%2B+f%5E2%29+-+e%29%2F2%29.

These values for r and s should be combined in such a way that the correct sign of their product r%2As+=+b%2F2 is provided.
Note that the second complex square root is the complex number opposite to the first one: w%5B2%5D+=+-w%5B1%5D.

So, now you know everything to calculate complex roots of the quadratic equation with the complex coefficients.

Example 1
Solve the quadratic equation
x%5E2+%2B+1+=+0.

This is the famous quadratic equation, which has the roots x%5B1%5D+=+red%28i%29 and x%5B2%5D+=+-red%28i%29
(see lessons Complex numbers and arithmetical operations over them and
How to take a square root of a complex number, Example 2) of this module.
I want to show you how the entire procedure works in this case.

Solution
We have a=1, b=0 and c=1.
The discriminant (9) is equal to
d+=+b%5E2+-4ac+=+0%5E2+-+4%2A1%2A1+=+-4.
Now, calculate the square root of the discriminant.
Presenting the discriminant as the complex number
d+=+e+%2B+red%28i%29%2Af
we have e+=+-4, f+=+0.
Therefore, the first value of the square root of the discriminant is the complex number
w%5B1%5D+=+r%5B1%5D+%2B+red%28i%29%2As%5B1%5D,
where
r%5B1%5D+=+sqrt%28%28-4+%2B+sqrt%28%28-4%29%5E2+%2B+0%5E2%29%29%2F2%29+=+sqrt%28%28-4+%2B+4%29%2F2%29+=+0, s%5B1%5D+=+sqrt%28%28sqrt%28%28-4%29%5E2+%2B+0%5E2%29+-+%28-4%29%29%2F2%29+=+sqrt%28%284+-+%28-4%29%29%2F2%29+=+2,
so w%5B1%5D+=+0+%2B+2%2Ared%28i%29+=+2%2Ared%28i%29.
The second value of the square root of the discriminant is the opposite complex number w%5B2%5D+=+0+-+2%2Ared%28i%29+=+-2%2Ared%28i%29.
Now, from the quadratic formula (8), the first root of the equation is
x%5B1%5D+=+%28b+%2B+sqrt%28d%29%29%2F%282a%29+=+%280+%2B+2%2Ared%28i%29%29%2F2+=+red%28i%29,
and the second root of the equation is
x%5B2%5D+=+%28b+-+sqrt%28d%29%29%2F%282a%29+=+%280+-+2%2Ared%28i%29%29%2F2+=+-red%28i%29,
as expected.

Example 2
Solve the quadratic equation in complex numbers
z%5E2+-+8%281-red%28i%29%29+%2B+63+-+16%2Ared%28i%29+=+0.

Solution
We have a=1, b=-8%2A%281-red%28i%29%29 and c=63-16%2Ared%28i%29.
The discriminant (9) is equal to
d+=+b%5E2+-4ac+=+64%2A%281-red%28i%29%29%5E2+-+4%2A1%2A%2863-16%2Ared%28i%29%29+=+-2%2A64%2Ared%28i%29+%2B64%2Ared%28i%29+-4%2A63+=+-4%2A63+-+64%2Ared%28i%29+=+-252+-64%2Ared%28i%29.
Now, calculate the square root of the discriminant.
Presenting the discriminant as the complex number
d+=+e+%2B+red%28i%29%2Af
we have e+=+-252, f+=+-64.
Therefore, the first value of the square root of the discriminant is the complex number
w%5B1%5D+=+r%5B1%5D+%2B+red%28i%29%2As%5B1%5D,
where
r%5B1%5D+=+sqrt%28%28-252+%2B+sqrt%28%28-252%29%5E2+%2B+%28-64%29%5E2%29%29%2F2%29+=+sqrt%28%28-252+%2B+260%29%2F2%29+=+sqrt%284%29+=+2, s%5B1%5D+=+sqrt%28%28sqrt%28%28-252%29%5E2+%2B+%28-64%29%5E2%29+-+%28-252%29%29%2F2%29+=+sqrt%28%28260+%2B+252%29%2F2%29+=+sqrt%28256%29+=+-16.
so w%5B1%5D+=+2+-+16%2Ared%28i%29.
The second value of the square root of the discriminant is the opposite complex number w%5B2%5D+=+-2+%2B+16%2Ared%28i%29.
Now, from the quadratic formula (8), the first root of the equation is
x%5B1%5D+=+%288-8%2Ared%28i%29+%2B+sqrt%28d%29%29%2F%282a%29+=+%288-8%2Ared%28i%29+%2B+w%5B1%5D%29%2F%282a%29+=+%288-8%2Ared%28i%29+%2B+2+-+16%2Ared%28i%29%29%2F2+=+%2810+-24%2Ared%28i%29%29%2F2+=+5+-+12%2Ared%28i%29,
and the second root of the equation is
x%5B2%5D+=+%288-8%2Ared%28i%29+-+sqrt%28d%29%29%2F%282a%29+=+%288-8%2Ared%28i%29+%2B+w%5B2%5D%29%2F%282a%29+=+%288-8%2Ared%28i%29+%2B%28-2%29+%2B+16%2Ared%28i%29%29%2F2+=+%286+%2B8%2Ared%28i%29%29%2F2+=+3+%2B+4%2Ared%28i%29.

Answer. The roots of the given quadratic equations are x%5B1%5D+=+5+-+12%2Ared%28i%29 and x%5B1%5D+=+3+%2B+4%2Ared%28i%29.



For your convenience, below is the list of my relevant lessons on complex numbers in this site in the logical order. They all are under the current topic Complex numbers in the section Algebra II.
Complex numbers and arithmetical operations over them
Complex plane
Addition and subtraction of complex numbers in complex plane
Multiplication and division of complex numbers in complex plane
Raising a complex number to an integer power
How to take a root of a complex number
Solution of the quadratic equation with real coefficients on complex domain
How to take a square root of a complex number
Solution of the quadratic equation with complex coefficients on complex domain  (this lesson)




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