Complex numbers

You saw that the square root of a negative number is not a real number. We call such a number an imaginary number.

An imaginary number is written in the form $b\,i$, where $b$ is a real number and $i=\sqrt{-1}$.

Since $i=\sqrt{-1}$, it follows that $i^2=-1$.

Writing the square root of a negative number in terms of $i$

We can write the square root of any negative number in terms of $i$. For example,

$\sqrt{-10}=\sqrt{-1}\cdot \sqrt{10}=i\sqrt{10}$

Example 1: Write in terms of $i$ and simplify.

(a) $\sqrt{-64}$ (b) $\sqrt{-50}$ (c) $-\sqrt{-108}$

Solution: Recall the perfect squares: $1, 4, 9, 16, 25, 36, 49, 64, 81, 100, \dots$

(a) $\sqrt{-64}=i\sqrt{64}=8i$

(b) $\begin{align*}\sqrt{-50}&=i\sqrt{50}\\&=i\sqrt{25\cdot 2}\\&=5i\sqrt{2}\end{align*}$

Example 2: Multiply.

(a) $\sqrt{-7}\cdot \sqrt{14}$ (b) $\sqrt{-16}\cdot \sqrt{-49}$ (c) $\sqrt{-6}\cdot \sqrt{-2}$

Solution: First write each square root of a negative number in terms of $i$, then use $i^2=-1$.

(a) $\begin{align*}\sqrt{-7}\cdot \sqrt{14}&=i\sqrt{7}\cdot \sqrt{14}\\&=i\sqrt{98}\\&=i\sqrt{49\cdot 2}\\&=7i\sqrt{2}\end{align*}$

(b) $\begin{align*}\sqrt{-16}\cdot \sqrt{-49}&=i\sqrt{16}\cdot i\sqrt{49}\\&=i^2\cdot 4\cdot 7\\&=(-1)(28)\\&=-28\end{align*}$

Note: You cannot write $\sqrt{-6}\cdot \sqrt{-2}=\sqrt{(-6)(-2)}=\sqrt{12}$. You must convert each factor to $i$ first.

(c) $\begin{align*}\sqrt{-6}\cdot \sqrt{-2}&=i\sqrt{6}\cdot i\sqrt{2}\\&=i^2\sqrt{12}\\&=-\sqrt{4\cdot 3}\\&=-2\sqrt{3}\end{align*}$

Example 3: Divide.

(a) $\dfrac{\sqrt{-294}}{\sqrt{-6}}$ (b) $\dfrac{\sqrt{-25}}{\sqrt{5}}$

Solution:

(a) $\begin{align*}\dfrac{\sqrt{-294}}{\sqrt{-6}}&=\dfrac{i\sqrt{294}}{i\sqrt{6}}\\&=\sqrt{\dfrac{294}{6}}\\&=\sqrt{49}\\&=7\end{align*}$

(b) $\begin{align*}\dfrac{\sqrt{-25}}{\sqrt{5}}&=\dfrac{i\sqrt{25}}{\sqrt{5}}\\&=i\sqrt{\dfrac{25}{5}}\\&=i\sqrt{5}\end{align*}$

Powers of $i$

The powers of $i$ follow a repeating pattern. Using $i^2=-1$,

$\begin{align*}i^2&=-1\\i^3&=i^2\cdot i=(-1)i=-i\\i^4&=i^2\cdot i^2=(-1)(-1)=1\end{align*}$

Since $i^4=1$, any power of $i$ whose exponent is a multiple of $4$ equals $1$. For example, $i^8=(i^4)^2=1$ and $i^{12}=(i^4)^3=1$.

To simplify a power of $i$, divide the exponent by $4$ and use the remainder.

Example 1: Simplify.

(a) $i^{77}$ (b) $i^{43}$ (c) $i^{98}$

Solution:

(a) $77\div 4$ leaves a remainder of $1$, so $i^{77}=i^{76}\cdot i=(i^4)^{19}\cdot i=1\cdot i=i$

(b) $43\div 4$ leaves a remainder of $3$, so $i^{43}=i^{40}\cdot i^3=1\cdot(-i)=-i$

Example 2: Simplify.

(a) $i^{-86}$ (b) $i^{-63}$

Solution: A negative exponent means take the reciprocal.

(a) $\begin{align*}i^{-86}&=\dfrac{1}{i^{86}}=\dfrac{1}{i^{84}\cdot i^{2}}\\&=\dfrac{1}{1\cdot(-1)}\\&=-1\end{align*}$

(b) $\begin{align*}i^{-63}&=\dfrac{1}{i^{63}}=\dfrac{1}{i^{60}\cdot i^{3}}\\&=\dfrac{1}{1\cdot(-i)}=\dfrac{1}{-i}\\&=\dfrac{1}{-i}\cdot \dfrac{i}{i}=\dfrac{i}{-i^2}\\&=\dfrac{i}{1}=i\end{align*}$

Complex numbers

The sum of a real number and an imaginary number is called a complex number. For example, $2+3i$ is a complex number: $2$ is the real part and $3$ is the imaginary part.

The standard form of a complex number is

$a+bi$

where $a$ is the real part and $b$ is the imaginary part.

Examples of complex numbers in standard form: $-6+5i$, $\dfrac{6}{7}-\dfrac{4}{11}i$, $-7-4i$, $-\dfrac{8}{3}+\dfrac{1}{2}i$.

Addition and subtraction of complex numbers

To add or subtract complex numbers, combine the real parts and combine the imaginary parts.

Example 1: Add or subtract. Write your answer in the standard form of a complex number.

(a) $(-4+11i)+(7-10i)$ (b) $(2+9i)-(-5+6i)$

Solution:

(a) $\begin{align*}(-4+11i)+(7-10i)&=(-4+7)+(11-10)i\\&=3+i\end{align*}$

(b) $\begin{align*}(2+9i)-(-5+6i)&=2+9i+5-6i\\&=(2+5)+(9-6)i\\&=7+3i\end{align*}$

Example 2: Add or subtract. Write your answer in the standard form of a complex number.

$(6-8i)-(-6+13i)+(5-4i)$

Solution:

$\begin{align*}(6-8i)-(-6+13i)+(5-4i)&=6-8i+6-13i+5-4i\\&=(6+6+5)+(-8-13-4)i\\&=17-25i\end{align*}$

Multiplication of complex numbers

To multiply complex numbers, use the distributive property, replace $i^2$ with $-1$, and write the result in standard form.

Example 1: Multiply. Write your answer in the standard form of a complex number.

$4i(2-12i)$

Solution:

$\begin{align*}4i(2-12i)&=8i-48i^2\\&=8i-48(-1)\\&=8i+48\\&=48+8i\end{align*}$

Example 2: Multiply. Write your answer in the standard form of a complex number.

$(1-8i)(-6-5i)$

Solution:

$\begin{align*}(1-8i)(-6-5i)&=-6-5i+48i+40i^2\\&=-6+43i+40(-1)\\&=-6+43i-40\\&=-46+43i\end{align*}$

Example 3: Multiply. Write your answer in the standard form of a complex number.

$-3i(4+7i)^2$

Solution: Use the formula $(A+B)^2=A^2+2AB+B^2$ to expand $(4+7i)^2$:

$\begin{align*}(4+7i)^2&=4^2+2\cdot 4\cdot 7i+(7i)^2\\&=16+56i+49i^2\\&=16+56i-49\\&=-33+56i\end{align*}$

Now multiply by $-3i$:

$\begin{align*}-3i(-33+56i)&=99i-168i^2\\&=99i+168\\&=168+99i\end{align*}$

Complex conjugate

The complex conjugate of $a+bi$ is $a-bi$. To form the conjugate, change the sign of the imaginary part.

For example, the conjugate of $8+12i$ is $8-12i$, and the conjugate of $10-7i$ is $10+7i$.

When you multiply a complex number by its conjugate, the result is a real number:

$\begin{align*}(a+bi)(a-bi)&=a^2-(bi)^2\\&=a^2-b^2i^2\\&=a^2-b^2(-1)\\&=a^2+b^2\end{align*}$

Division of complex numbers

To divide complex numbers, multiply the numerator and denominator by the conjugate of the denominator. This makes the denominator a real number.

Example 1: Divide. Write your answer in the standard form of a complex number.

$\dfrac{25i}{3+4i}$

Solution: Multiply the numerator and denominator by the conjugate of the denominator, $3-4i$:

$\begin{align*}\dfrac{25i}{3+4i}&=\dfrac{25i}{3+4i}\cdot \dfrac{3-4i}{3-4i}\\&=\dfrac{25i(3-4i)}{3^2+4^2}\\&=\dfrac{75i-100i^2}{9+16}\\&=\dfrac{75i+100}{25}\\&=\dfrac{100}{25}+\dfrac{75}{25}i\\&=4+3i\end{align*}$

Example 2: Divide. Write your answer in the standard form of a complex number.

$\dfrac{5+3i}{7-4i}$

Solution: Multiply the numerator and denominator by the conjugate $7+4i$:

$\begin{align*}\dfrac{5+3i}{7-4i}&=\dfrac{(5+3i)(7+4i)}{7^2+4^2}\\&=\dfrac{35+20i+21i+12i^2}{49+16}\\&=\dfrac{35+41i-12}{65}\\&=\dfrac{23+41i}{65}\\&=\dfrac{23}{65}+\dfrac{41}{65}i\end{align*}$

Example 3: Divide. Write your answer in the standard form of a complex number.

$\dfrac{8-5i}{2i}$

Solution: Multiply the numerator and denominator by $-i$:

$\begin{align*}\dfrac{8-5i}{2i}&=\dfrac{(8-5i)(-i)}{(2i)(-i)}\\&=\dfrac{-8i+5i^2}{-2i^2}\\&=\dfrac{-8i-5}{2}\\&=-\dfrac{5}{2}-4i\end{align*}$

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