Complex numbers
The equation ${x^2=-1}$ has no real solution, because the square of a real number is never negative. On the page on quadratic equations, we saw that some quadratic equations have no real solutions for the same reason.
To solve every quadratic equation, mathematicians made the number system bigger. They added a new number, ${i}$, whose square is ${-1}$. The result is the system of complex numbers.
The imaginary unit
The imaginary unit ${i}$ is the number with
${i^2=-1}$, that is, ${i=\sqrt{-1}}$.
A complex number is a number of the form
${a+bi}$,
where ${a}$ and ${b}$ are real numbers. The number ${a}$ is the real part, and ${b}$ is the imaginary part. For example, ${3-4i}$ has real part ${3}$ and imaginary part ${-4}$.
- If ${b=0}$, the number ${a+0i=a}$ is a real number. So every real number is also a complex number.
- If ${a=0}$ and ${b\ne 0}$, the number ${bi}$ is called pure imaginary, like ${5i}$.
Two complex numbers are equal when their real parts are equal and their imaginary parts are equal.
The word "imaginary" is only a name. Complex numbers are used every day in electrical engineering and physics.
Adding and subtracting
Add or subtract the real parts, and add or subtract the imaginary parts. Treat ${i}$ like a variable:
${(a+bi)+(c+di)=(a+c)+(b+d)i}$
Example 1: Write each answer in the form ${a+bi}$.
(a) ${(3+2i)+(5-7i)}$
(b) ${(4-i)-(-2+3i)}$
Solution:
(a) Combine the real parts and the imaginary parts:
${(3+5)+(2-7)i=8-5i}$
(b) The minus sign changes the sign of both parts of ${-2+3i}$:
$\begin{align*}&4-i+2-3i\\&=(4+2)+(-1-3)i\\&=6-4i\end{align*}$
Multiplying
Multiply complex numbers like binomials (with FOIL). Then replace ${i^2}$ by ${-1}$, and combine the real parts.
Example 2: Write each product in the form ${a+bi}$.
(a) ${(2+3i)(4-5i)}$
(b) ${(3+i)^2}$
(c) ${(2+5i)(2-5i)}$
Solution:
(a) Use FOIL. Then use ${i^2=-1}$, so ${-15i^2=+15}$:
$\begin{align*}&(2+3i)(4-5i)\\&=8-10i+12i-15i^2\\&=8+2i+15\\&=23+2i\end{align*}$
(b) Use ${(a+b)^2=a^2+2ab+b^2}$:
$\begin{align*}&(3+i)^2\\&=9+6i+i^2\\&=9+6i-1\\&=8+6i\end{align*}$
(c) Use ${(a+b)(a-b)=a^2-b^2}$:
$\begin{align*}&(2+5i)(2-5i)\\&=4-25i^2\\&=4+25\\&=29\end{align*}$
The answer to (c) is a real number. That is not an accident, as we see next.
Conjugates
The complex conjugate of ${a+bi}$ is ${a-bi}$: the same number with the sign of the imaginary part changed. A number times its conjugate is always a real number:
${(a+bi)(a-bi)=a^2+b^2}$
This is what makes division possible.
Dividing
To divide by a complex number, multiply the numerator and the denominator by the conjugate of the denominator. The denominator becomes a real number. This is the same idea as rationalizing a denominator.
Example 3: Write each quotient in the form ${a+bi}$.
(a) ${\dfrac{3+2i}{1-i}}$
(b) ${\dfrac{4-i}{i}}$
Solution:
(a) The conjugate of ${1-i}$ is ${1+i}$. Multiply the numerator and the denominator by it:
$\begin{align*}&\dfrac{3+2i}{1-i}\cdot\dfrac{1+i}{1+i}\\&=\dfrac{3+3i+2i+2i^2}{1^2+1^2}\\&=\dfrac{3+5i-2}{2}\\&=\dfrac{1+5i}{2}\end{align*}$
Split the fraction into its real and imaginary parts:
$\dfrac{1+5i}{2}=\dfrac{1}{2}+\dfrac{5}{2}i$
(b) The denominator ${i=0+1i}$ has conjugate ${-i}$. Multiply the numerator and the denominator by ${-i}$:
$\begin{align*}&\dfrac{4-i}{i}\cdot\dfrac{-i}{-i}\\&=\dfrac{-4i+i^2}{-i^2}\\&=\dfrac{-4i-1}{1}\\&=-1-4i\end{align*}$
Powers of ${i}$
The powers of ${i}$ repeat in a cycle of four:
| Power | Value |
|---|---|
| ${i^1}$ | ${i}$ |
| ${i^2}$ | ${-1}$ |
| ${i^3=i^2\cdot i}$ | ${-i}$ |
| ${i^4=i^2\cdot i^2}$ | ${1}$ |
After ${i^4=1}$, the cycle starts again: ${i^5=i}$, ${i^6=-1}$, and so on. To find a large power, divide the exponent by ${4}$. Only the remainder matters.
Example 4: Find ${i^{23}}$.
Solution:
${23=4\cdot 5+3}$, so the remainder is ${3}$. Use ${i^4=1}$:
$\begin{align*}&i^{23}\\&=\left(i^4\right)^5\cdot i^3\\&=1^5\cdot(-i)\\&=-i\end{align*}$
Square roots of negative numbers
If ${a>0}$, the principal square root of ${-a}$ is
${\sqrt{-a}=i\sqrt{a}}$.
For example, ${\sqrt{-9}=3i}$ and ${\sqrt{-5}=i\sqrt{5}}$. Check: ${(3i)^2=9i^2=-9}$.
Watch out: always write a negative radicand in terms of ${i}$ before you multiply. The rule ${\sqrt{a}\,\sqrt{b}=\sqrt{ab}}$ does not work when both ${a}$ and ${b}$ are negative.
Example 5: Find ${\sqrt{-4}\cdot\sqrt{-9}}$.
Solution:
Write each root in terms of ${i}$ first:
$\begin{align*}&\sqrt{-4}\cdot\sqrt{-9}\\&=2i\cdot 3i\\&=6i^2\\&=-6\end{align*}$
The wrong way gives ${\sqrt{(-4)(-9)}=\sqrt{36}=6}$. The sign is wrong.
Quadratic equations with complex solutions
When the discriminant ${b^2-4ac}$ is negative, the quadratic formula gives two complex solutions. They are always conjugates of each other: ${p+qi}$ and ${p-qi}$.
Example 6: Solve each equation.
(a) ${x^2-4x+13=0}$
(b) ${x^2+x+1=0}$
Solution:
(a) Here ${a=1}$, ${b=-4}$, ${c=13}$. Use the quadratic formula:
$\begin{align*}&x\\&=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(1)(13)}}{2(1)}\\&=\dfrac{4\pm\sqrt{16-52}}{2}\\&=\dfrac{4\pm\sqrt{-36}}{2}\end{align*}$
Write ${\sqrt{-36}=6i}$. Then divide each term by ${2}$:
$\begin{align*}&x\\&=\dfrac{4\pm 6i}{2}\\&=2\pm 3i\end{align*}$
Check ${x=2+3i}$. First, ${(2+3i)^2=4+12i+9i^2=-5+12i}$. Put it into the equation:
$\begin{align*}&(2+3i)^2-4(2+3i)+13\\&=(-5+12i)-8-12i+13\\&=0\end{align*}$
(b) This equation came up on the page on quadratic equations, where its discriminant was ${-3}$. Here ${a=1}$, ${b=1}$, ${c=1}$:
$\begin{align*}&x\\&=\dfrac{-1\pm\sqrt{1-4}}{2}\\&=\dfrac{-1\pm\sqrt{-3}}{2}\\&=\dfrac{-1\pm i\sqrt{3}}{2}\end{align*}$
In the form ${a+bi}$, the solutions are ${-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i}$ and ${-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i}$.
Summary
- ${i^2=-1}$. A complex number is ${a+bi}$, with real part ${a}$ and imaginary part ${b}$.
- Add and subtract the real parts and the imaginary parts separately.
- Multiply like binomials, and replace ${i^2}$ by ${-1}$.
- The conjugate of ${a+bi}$ is ${a-bi}$, and ${(a+bi)(a-bi)=a^2+b^2}$.
- To divide, multiply the numerator and the denominator by the conjugate of the denominator.
- Powers of ${i}$ repeat every four: ${i,\ -1,\ -i,\ 1}$.
- ${\sqrt{-a}=i\sqrt{a}}$ for ${a>0}$. Write roots of negatives with ${i}$ before multiplying.
- When ${b^2-4ac<0}$, a quadratic equation has two complex conjugate solutions.