# Operations with Complex Numbers

### Adding and Subtracting Complex Numbers Complex numbers are added and subtracted by combining like terms.

A term of an algebraic expression is a part of the expression that is separated from other parts by addition or subtraction. Like terms, such as $3x$ and $-8x$, are terms that contain the same variable (or variables) with the same exponents. Adding and subtracting complex numbers is very similar to adding and subtracting algebraic expressions by combining like terms.

• For subtraction, distribute the negative, and then add.

The letter $z$ is often used to represent a complex number. If an expression or equation includes two complex numbers, $z_1$ can represent the first complex number, and $z_2$ can represent the second complex number.

Step-By-Step Example
Equations for $z_1$ and $z_2$ are given:
\begin{aligned}z_{1}&=12+2i\\z_{2}&=8-4i\end{aligned}
Evaluate each of these expressions:

1) $z_1+z_2$

2) $z_1-z_2$

Step 1

Write the addition and subtraction expressions. For subtraction, distribute the minus sign.

For $z_1+z_2$:
$\begin{gathered}z_1+z_2\\(12+2i)+(8-4i)\end{gathered}$
For $z_1-z_2$:
$\begin{gathered}z_1-z_2\\(12+2i)-(8-4i)\\12+2i-8+4i\end{gathered}$
Step 2

Reorder so that like terms are next to each other.

For $z_1+z_2$:
$\begin{gathered}(12+2i)+(8-4i)\\12+2i+8-4i\\12+8+2i-4i\end{gathered}$
For $z_1-z_2$:
$\begin{gathered}12+2i-8+4i\\12-8+2i+4i\end{gathered}$
Solution

Combine like terms.

For $z_1+z_2$:
$\begin{gathered}12+8+2i-4i\\20-2i\end{gathered}$
For $z_1-z_2$:
$\begin{gathered}12-8+2i+4i\\4+6i\end{gathered}$

### Multiplying Complex Numbers Powers of $i$ are computed using the property $i^2 = -1$. Complex numbers are multiplied by using the distributive property and powers of $i$.

### Powers of i

It is important to remember that $i$ is not a variable; it has a value of $\sqrt{-1}$. So, unlike powers of variables, such as $x^2$, powers of $i$ need to be simplified. When calculating powers of $i$ starting from $i^1$ and onward, a pattern emerges. Analyzing the pattern involves two rules of exponents: the product of powers property and the power of a power property.

The product of powers property states that two powers with the same base can be multiplied by adding their exponents. For example:
\begin{aligned}i^{4}\cdot i^{2}&=i^{4+2}\\&=i^6\end{aligned}
The property can also be applied to rewrite a single power as a product of two powers with the same base:
\begin{aligned}i^{4}\cdot i^{2}&=i^{4+2}\\&=i^6\end{aligned}
The power of a power property states that a power can be raised to a power by multiplying the exponents. For instance:
\begin{aligned}\left(i^{4}\right)^{2}&=i^{4\cdot 2}\\&=i^8\end{aligned}
The property can be also applied to rewrite a single power as a power of a power:
\begin{aligned}i^{8}&=i^{4\cdot 2}\\&=\left(i^{4}\right)^{2}\end{aligned}

### Simplifying Powers of $i$

Power of $i$ Expanded Form Simplified Form
$i^1$
$i^1=i$
$i$
$i^2$
\begin{aligned}i^2&=\left(\sqrt{-1}\right)^{2}\\&=-1\end{aligned}
$-1$
$i^3$
\begin{aligned}i^3 &= i^2 \cdot i\\&=-1\cdot i\\&=-i\end{aligned}
$-i$
$i^4$
\begin{aligned}i^4&=i^2 \cdot i^2 \\&= -1 \cdot (-1)\\&=1\end{aligned}
$1$
$i^5$
\begin{aligned}i^5 &= i^4 \cdot i\\&=1\cdot i\\&=i\end{aligned}
$i$
$i^6$
\begin{aligned}i^6 &= i^4 \cdot i^2\\&=1\cdot i^2\\&=-1\end{aligned}
$-1$
$i^7$
\begin{aligned}i^7 &= i^4 \cdot i^3\\&=1\cdot i^3\\&=-i\end{aligned}
$-i$
$i^8$
\begin{aligned}i^8 &= i^4 \cdot i^4\\&=1\cdot i^4\\&=1\end{aligned}
$1$
$i^9$
\begin{aligned}i^9 &= \left(i^{4}\right)^{2} \cdot i\\&=1^{2} \cdot i\\&=i\end{aligned}
$i$

There are only four possible values for a power of $i$, and they repeat in the pattern: $i$, -1, $-i$, 1.

Analyzing the calculations of the powers of $i$ shows that for every power greater than $i^4$, it is possible to factor out a power of $i^4$ and write it as 1.

Step-By-Step Example
Simplify Powers of $i$
Simplify $i^{14}$.
Step 1
Apply the product of powers property to factor into two powers. One exponent should be a multiple of 4.
\begin{aligned}i^{14}&=i^{12+2}\\&=i^{12} \cdot i^2\end{aligned}
Step 2
Apply the power of a power property to rewrite the first factor as a power of $i^4$.
\begin{aligned}i^{12} \cdot i^2&=i^{4\cdot 3}\cdot i^{2}\\&=\left(i^4\right)^3 \cdot i^2\end{aligned}
Step 3
Write $i^4$ as 1.
$\left(i^{4}\right)^{3} \cdot i^{2}=\left(1\right)^{3} \cdot i^{2}$
Solution
Simplify.
\begin{aligned}\left(1\right)^3 \cdot i^2&=1 \cdot\left(-1\right)\\&=-1\end{aligned}
So, $i^{14}$ simplifies to:
$i^{14}=-1$

### Products of Complex Numbers

The product property of square roots states that for real numbers $a$ and $b$, if $a\geq0$ and $b\geq0$, then:
$\sqrt{a} \cdot \sqrt{b}=\sqrt{ab}$
However, not all properties of real numbers associated with square roots carry over to complex numbers. If $a$ and $b$ are negative numbers:
$\sqrt{a} \cdot \sqrt{b}\neq \sqrt{ab}$
For example:
\begin{aligned}i^2&=\sqrt{-1} \cdot \sqrt{-1}\\i^2&\stackrel{?}{=}\sqrt{-1\cdot (-1)}\\i^2&\stackrel{?}{=}\sqrt1\\i^2&\neq1\end{aligned}
Applying the product property of square roots leads to a false statement. The value of $i^2$ is -1, not 1. So, the result is:
$\sqrt{-1}\cdot\sqrt{-1}\neq\sqrt{-1\cdot (-1)}$
Therefore, the product property of square roots does not apply to the square roots of negative numbers. For this reason, when multiplying complex numbers, it is important to first write them in the standard form of a complex number:
$a+bi$
Next, use the distributive property to multiply and then simplify the result. When using the distributive property to multiply complex numbers, multiply as with binomials, or polynomials with two terms. One way to multiply two binomials, such as $x+3$ and $2x-5$, is to multiply the first terms, the outside terms, the inside terms, and the last terms (FOIL), and then add the products.
\begin{aligned}(x+3)(2x-5)&=2x^2-5x+6x-15\\&=2x^2+x-15\end{aligned}
Step-By-Step Example
Multiply Complex Numbers
Multiply the expression:
$(\sqrt{-4}+6)(1+\sqrt{-16})$
Step 1
Write the first factor in the standard form of a complex number, or $a+bi$.
$\begin{gathered}\sqrt{-4}+6\\6+2i\end{gathered}$
Then write the second factor in the standard form of a complex number.
$\begin{gathered}1+\sqrt{-16}\\1+\sqrt{-1 \cdot 16}\\1+4i\end{gathered}$
Put the factors together:
$\begin{gathered}(\sqrt{-4}+6)(1+\sqrt{-16})\\(6+2i)(1+4i)\end{gathered}$
Step 2
Use the distributive property (FOIL) to multiply the two numbers.
$\begin{gathered}(6+2i)(1+4i)\\6+24i+2i+8i^2\end{gathered}$
Step 3
Write $i^2$ as –1.
$\begin{gathered}6+24i+2i+8i^2\\6+24i+2i+8(-1)\\6+24i+2i-8\end{gathered}$
Solution
Combine like terms, and write them in the standard form of a complex number.
$\begin{gathered}6+24i+2i-8\\-2+26i\end{gathered}$
For a given complex number $a+bi$, the complex conjugate is the number with the same real part and the opposite imaginary part, $a-bi$. For $z=a+bi$, the complex conjugate is written as $\bar z=a-bi$. The product of a complex number and its complex conjugate, or $z\bar{z}$, is a real number:
\begin{aligned}z\bar{z}&=(a+bi)(a-bi)\\&=a^2 +abi-abi-b^2i^2\\&=a^2-b^2(-1)\\&=a^2+b^2\end{aligned}
Step-By-Step Example
Multiply a Complex Number by Its Complex Conjugate
Determine $z\bar z$ for the equation:
$z=7+2i$
Step 1
Determine the complex conjugate by changing the sign of the imaginary part.
$\bar z=7-2i$
Step 2
Multiply the complex number and its complex conjugate.
$z\bar z=(7+2i)(7-2i)$
Solution
Simplify the product.
\begin{aligned}z\bar{z}&=49-14i+14i-4i^2\\&=49-(4)(-1)\\&=49-(-4)\\&=49+4\\&=53\end{aligned}
Check that $z\bar {z}=a^2+b^2$, where $z=7+2i$, $a=7$, and $b=2$.
\begin{aligned}a^2+b^2&=\left(7\right)^{2}+\left(2\right)^{2}\\&=49+4\\&=53\end{aligned}

### Dividing Complex Numbers Division by a complex number $a + bi$ is performed by multiplying the numerator and denominator by the complex conjugate, $a - bi$.
A radical expression is an expression that contains at least one radical sign, $\sqrt{\phantom{0}}$. Rationalizing the denominator of a fraction is the process of rewriting the fraction without a radical expression in the denominator. A denominator with two terms can be rationalized by multiplying the numerator and denominator of the fraction by the conjugate of the denominator. For the radical expression $2+\sqrt{3}$, the conjugate is $2-\sqrt{3}$. The product of a radical expression and its conjugate is a rational number.
\begin{aligned}(2+\sqrt{3})(2-\sqrt{3})&=2^{2}-2\sqrt{3}+2\sqrt{3}-(\sqrt{3})^2\\&=4-2\sqrt{3}+2\sqrt{3}-3\\&=4-3\\&=1\end{aligned}
The process of dividing complex numbers is similar to rationalizing the denominator because it uses the fact that the product of a complex number and its complex conjugate is a real number. To divide a complex number $z_1$ by a complex number $z_2$:
1. Write the two numbers as a fraction:
$\frac{z_1}{z_2}$
2. Multiply the fraction by this fraction, which is equivalent to 1:
$\frac{\bar z_2}{\bar z_2}$
3. Simplify the result. The denominator will be a real number.
$\frac{z_1 \bar z_2}{z_2 \bar z_2}$
4. Write it in the standard form of a complex number.
Step-By-Step Example
Divide Complex Numbers
Divide $6-2i$ by $1+2i$.
Step 1
Write $6-2i$ as the numerator of a fraction and $1+2i$ as the denominator.
$\frac {6-2i}{1+2i}$
Step 2
The complex conjugate of $1+2i$ is $1-2i$. Make the complex conjugate a fraction that is equivalent to 1:
$\frac{1-2i}{1-2i}$
Multiply the fraction by the fraction of the complex conjugate.
$\frac {6-2i}{1+2i}=\frac {6-2i}{1+2i} \cdot \frac {1-2i}{1-2i}$
Step 3
Simplify the product.
\begin{aligned}\frac{6-2i}{1+2i} \cdot \frac{1-2i}{1-2i}&=\frac{6-12i-2i+4i^2}{1-2i+2i-4i^2}\\&=\frac{6-14i+4(-1)}{1-4(-1)}\\&=\frac{6-14i-4}{1+4}\\&=\frac{2-14i}{5}\end{aligned}
Solution
Write in the standard form of a complex number.
$\frac{2-14i}{5}=\frac{2}{5}-\frac{14}{5}i$