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Section CNO Complex Number Operations

In this section we review some of the basics of working with complex numbers.

Subsection CNA Arithmetic with complex numbers

A complex number is a linear combination of \(1\) and \(i=\sqrt{-1}\text{,}\) typically written in the form \(a+bi\text{.}\) Complex numbers can be added, subtracted, multiplied and divided, just like we are used to doing with real numbers, including the restriction on division by zero. We will not define these operations carefully immediately, but instead first illustrate with examples.

Example ACN. Arithmetic of complex numbers.

For example,
\begin{align*} (2+5i)+(6-4i)&=(2+6)+(5+(-4))i=8+i\\ (2+5i)-(6-4i)&=(2-6)+(5-(-4))i=-4+9i\\ (2+5i)(6-4i)&=(2)(6)+(5i)(6)+(2)(-4i)+(5i)(-4i)=12+30i-8i-20i^2\\ &=12+22i-20(-1)=32+22i\\ \end{align*}
Division takes just a bit more care. We multiply the denominator by a complex number chosen to produce a real number and then we can produce a complex number as a result.
\begin{align*} \frac{2+5i}{6-4i}&=\frac{2+5i}{6-4i}\frac{6+4i}{6+4i}=\frac{-8+38i}{52}=-\frac{8}{52}+\frac{38}{52}i=-\frac{2}{13}+\frac{19}{26}i\text{.} \end{align*}
In this example, we used \(6+4i\) to convert the denominator in the fraction to a real number. This number is known as the conjugate, which we define in the next section.
We will often exploit the basic properties of complex number addition, subtraction, multiplication and division, so we will carefully define the two basic operations, together with a definition of equality, and then collect nine basic properties in a theorem.

Definition CNE. Complex Number Equality.

The complex numbers \(\alpha=a+bi\) and \(\beta=c+di\) are equal, denoted \(\alpha=\beta\text{,}\) if \(a=c\) and \(b=d\text{.}\)

Definition CNA. Complex Number Addition.

The sum of the complex numbers \(\alpha=a+bi\) and \(\beta=c+di\) , denoted \(\alpha+\beta\text{,}\) is \((a+c)+(b+d)i\text{.}\)

Definition CNM. Complex Number Multiplication.

The product of the complex numbers \(\alpha=a+bi\) and \(\beta=c+di\) , denoted \(\alpha\beta\text{,}\) is \((ac-bd)+(ad+bc)i\text{.}\)

Proof.

We could derive each of these properties of complex numbers with a proof that builds on the identical properties of the real numbers. The only proof that might be at all interesting would be to show Property MICN since we would need to trot out a conjugate. For this property, and especially for the others, we might be tempted to construct proofs of the identical properties for the reals. This would take us way too far afield, so we will draw a line in the sand right here and just agree that these nine fundamental behaviors are true. OK?
Mostly we have stated these nine properties carefully so that we can make reference to them later in other proofs. So we will be linking back here often.
Zero and one play special roles, of course, and especially zero. Our first result is one we take for granted, but it requires a proof, derived from our nine properties. You can compare it to its vector space counterparts, Theorem ZSSM and Theorem ZVSM.

Proof.

We have
\begin{align*} 0\alpha&=0\alpha+0&& \knowl{./knowl/xref/property-ZCN.html}{\text{Property ZCN}}\\ &=0\alpha+\left(0\alpha-\left(0\alpha\right)\right)&& \knowl{./knowl/xref/property-AICN.html}{\text{Property AICN}}\\ &=\left(0\alpha+0\alpha\right)-\left(0\alpha\right)&& \knowl{./knowl/xref/property-AACN.html}{\text{Property AACN}}\\ &=\left(0+0\right)\alpha-\left(0\alpha\right)&& \knowl{./knowl/xref/property-DCN.html}{\text{Property DCN}}\\ &=0\alpha-\left(0\alpha\right)&& \knowl{./knowl/xref/property-ZCN.html}{\text{Property ZCN}}\\ &=0&& \knowl{./knowl/xref/property-AICN.html}{\text{Property AICN}}\text{.} \end{align*}
Our next theorem could be called “cancellation”, since it will make that possible. Though you will never see us drawing slashes through parts of products. We will also make very limited use of this result, or its vector space counterpart, Theorem SMEZV.

Proof.

(⇒) 
We conduct the forward argument in two cases. First suppose that \(\alpha=0\text{.}\) Then we are done. (That was easy.)
For the second case, suppose now that \(\alpha\neq 0\text{.}\) Then
\begin{align*} \beta&=1\beta&& \knowl{./knowl/xref/property-OCN.html}{\text{Property OCN}}\\ &=\left(\frac{1}{\alpha}\alpha\right)\beta&& \knowl{./knowl/xref/property-MICN.html}{\text{Property MICN}}\\ &=\frac{1}{\alpha}\left(\alpha\beta\right)&& \knowl{./knowl/xref/property-MACN.html}{\text{Property MACN}}\\ &=\frac{1}{\alpha}0&&\text{Hypothesis}\\ &=0&& \knowl{./knowl/xref/theorem-ZPCN.html}{\text{Theorem ZPCN}}\text{.} \end{align*}
(⇐) 
With two applications of Theorem ZPCN it is easy to see that if one of the scalars is zero, then so is the product.
As an equivalence (Proof Technique E), we could restate this result as the contrapositive (Proof Technique CP) by negating each statement, so it would read “\(\alpha\beta\neq 0\) if and only if \(\alpha\neq 0\) and \(\beta\neq 0\text{.}\)” After you have learned more about nonsingular matrices and matrix multiplication, you should compare this result with Theorem NPNF.

Subsection CCN Conjugates of Complex Numbers

Definition CCN. Conjugate of a Complex Number.

The conjugate of the complex number \(\alpha=a+bi\in\complexes\) is the complex number \(\conjugate{\alpha}=a-bi\text{.}\)

Example CSCN. Conjugate of some complex numbers.

For example,
\begin{align*} \conjugate{2+3i}=2-3i&&\conjugate{5-4i}=5+4i&& \conjugate{-3+0i}=-3+0i&&\conjugate{0+0i}=0+0i\text{.} \end{align*}
Notice how the conjugate of a real number leaves the number unchanged. The conjugate enjoys some basic properties that are useful when we work with linear expressions involving addition and multiplication.

Proof.

Let \(\alpha=a+bi\) and \(\beta=r+si\text{.}\) Then
\begin{equation*} \conjugate{\alpha+\beta}=\conjugate{(a+r)+(b+s)i}=(a+r)-(b+s)i=(a-bi)+(r-si)=\conjugate{\alpha}+\conjugate{\beta}\text{.} \end{equation*}

Proof.

Let \(\alpha=a+bi\) and \(\beta=r+si\text{.}\) Then
\begin{align*} \conjugate{\alpha\beta}&=\conjugate{(ar-bs)+(as+br)i}=(ar-bs)-(as+br)i\\ &=(ar-(-b)(-s))+(a(-s)+(-b)r)i=(a-bi)(r-si)=\conjugate{\alpha}\conjugate{\beta}\text{.} \end{align*}

Proof.

Let \(\alpha=a+bi\text{.}\) Then
\begin{equation*} \conjugate{\conjugate{\alpha}}=\conjugate{a-bi}=a-(-bi)=a+bi=\alpha\text{.} \end{equation*}

Subsection MCN Modulus of a Complex Number

We define one more operation with complex numbers that may be new to you.

Definition MCN. Modulus of a Complex Number.

The modulus of the complex number \(\alpha=a+bi\in\complexes\text{,}\) is the nonnegative real number
\begin{equation*} \modulus{\alpha}=\sqrt{\conjugate{\alpha}\alpha}=\sqrt{a^2+b^2}\text{.} \end{equation*}

Example MSCN. Modulus of some complex numbers.

For example,
\begin{align*} \modulus{2+3i}&=\sqrt{13}& \modulus{5-4i}&=\sqrt{41}& \modulus{-3+0i}&=3& \modulus{0+0i}&=0\text{.} \end{align*}
The modulus can be interpreted as a version of the absolute value for complex numbers, as is suggested by the notation employed. You can see this in how \(\modulus{-3}=\modulus{-3+0i}=3\text{.}\) Notice too how the modulus of the complex zero, \(0+0i\text{,}\) has value \(0\text{.}\)
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