How to Solve Quadratic Equations with Complex Numbers

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Encyclopedia>Numbers and Quantities>Numbers>Complex Numbers>Equations>How to Solve Quadratic Equations with Complex Numbers

When solving quadratic equations using the quadratic formula, you sometimes get a negative value under the square root. In these cases, the equation does not have any real solutions.

But now that you’re working with complex numbers, you’re able to find all the solutions to quadratic equations. The reason for this is the fact that the imaginary unit $i$ can be utilized to find complex solutions to the quadratic formula.

Rule

Complex Quadratic Equations

Let $a, b, c \in ℂ$ be complex numbers with $a \neq 0$ . Then $a z^{2} + b z + c = 0$ has the following solutions:

z = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} .

If the expression $b^{2} - 4 a c$ is negative, you have to use the imaginary unit $i$ to find the solutions.

Example 1

Solve $z^{2} - 4 z + 5 = 0$ for $z$

You recognize the coefficients of the equation to be $a = 1$ , $b = - 4$ and $c = 5$ . This means that you can use the quadratic formula to find the solutions:

\begin{array}{l} z & = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = \frac{- (- 4) \pm \sqrt{{(4)}^{2} - 4 \cdot 1 \cdot 5}}{2 \cdot 1} \\ = \frac{4 \pm \sqrt{- 4}}{2} . \end{array}

Since you have a negative number inside the square root, the equation has no real solutions. However, by utilizing the imaginary, you can still find complex solutions:

\begin{array}{l} \sqrt{- 4} & = \sqrt{(4) (- 1)} \\ = \sqrt{4} \sqrt{- 1} \\ = 2 i . \end{array}

\sqrt{- 4} = \sqrt{(4) (- 1)} = \sqrt{4} \sqrt{- 1} = 2 i .

Thus the final solutions to the equation are:

\begin{matrix} z = \frac{4 \pm \sqrt{- 4}}{2} = \frac{4 \pm 2 i}{2} = 2 \pm i \\ ⇓ \\ z_{1} = 2 + i and z_{2} = 2 - i . \end{matrix}

Note! In Example 1 you used the following relation:

\sqrt{- 4} = \sqrt{(4) (- 1)} = \sqrt{4} \sqrt{- 1} = 2 i .

This relation does not hold in general for complex numbers. Using this relation for complex numbers can yield inconsistencies. The following is an example of such inconsistencies:

\begin{array}{l} - 1 & = i \cdot i \\ = \sqrt{- 1} \cdot \sqrt{- 1} \\ = \sqrt{- 1 \cdot - 1} \\ = \sqrt{1} \\ = 1 . \end{array}

- 1 = i \cdot i = \sqrt{- 1} \cdot \sqrt{- 1} = \sqrt{- 1 \cdot - 1} = \sqrt{1} = 1 .

The reason this calculation yields a contradiction is that

\sqrt{- 1} \cdot \sqrt{- 1} \neq \sqrt{- 1 \cdot - 1}

. The rule stating that

\sqrt{a b} = \sqrt{a} \sqrt{b}

only holds when both

a

and

b

are positive numbers.

By using the quadratic formula, you can even solve quadratic equations involving complex coefficients. In those cases, you need some knowledge about complex roots in order to simplify the expression inside the square root.

Example 2

Solve $\frac{1}{2} z^{2} + i z + \frac{\sqrt{3}}{2} i = 0$ for $z$

You recognize the coefficients as $a = \frac{1}{2}$ , $b = i$ and $c = \frac{\sqrt{3}}{2} i$ . This means that you can use the quadratic formula to find the solutions:

\begin{array}{l} z & = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = \frac{- i \pm \sqrt{i^{2} - 4 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2}} i}{2 \cdot \frac{1}{2}} \\ = - i \pm \sqrt{- 1 - \sqrt{3} i} . \end{array}

All complex numbers $w$ have two square roots. In this case you only need to consider the root whose argument lies in the interval $[0, π)$ . The reason for this is that both solutions are included in $\pm \sqrt{w}$ . In order to find the square root of $w = - 1 - \sqrt{3} i$ , you first need to write $w$ in polar form. In this case, the norm of $w$ is $r = 2$ , while the argument is $𝜃 = \frac{4 π}{3}$ . Thus $w = 2 e^{i \frac{4 π}{3}}$ in polar form. You can now find the square root of $w$ by taking the square root of the norm of $w$ and dividing the argument of $w$ by $2$ . Thus the square root of $w$ is $\sqrt{2} e^{i \frac{2 π}{3}}$ . In Cartesian form, the square root is written as $- \frac{\sqrt{2}}{2} + \frac{\sqrt{6}}{2} i$ . Thus the solutions to the equation are:

\begin{matrix} z = - i \pm (- \frac{\sqrt{2}}{2} + \frac{\sqrt{6}}{2} i) \\ ⇓ \\ z_{1} = - \frac{\sqrt{2}}{2} + (\frac{\sqrt{6}}{2} - 1) i \end{matrix}

and

\begin{matrix} z_{2} = \frac{\sqrt{2}}{2} + (- \frac{\sqrt{6}}{2} - 1) i . \end{matrix}

\begin{matrix} z = - i \pm (- \frac{\sqrt{2}}{2} + \frac{\sqrt{6}}{2} i) \\ ⇓ \\ z_{1} = - \frac{\sqrt{2}}{2} + (\frac{\sqrt{6}}{2} - 1) i and z_{2} = \frac{\sqrt{2}}{2} + (- \frac{\sqrt{6}}{2} - 1) i . \end{matrix}

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