search
Search
Unlock 100+ guides
search toc
close
account_circle
Profile
exit_to_app
Sign out
What does this mean?
Why is this true?
Give me some examples!
search
keyboard_voice
close
Searching Tips
Search for a recipe:
"Creating a table in MySQL"
Search for an API documentation: "@append"
Search for code: "!dataframe"
Apply a tag filter: "#python"
Useful Shortcuts
/ to open search panel
Esc to close search panel
to navigate between search results
d to clear all current filters
Enter to expand content preview
Doc Search
Code Search Beta
SORRY NOTHING FOUND!
mic
Start speaking...
Voice search is only supported in Safari and Chrome.
Shrink
Navigate to
function
Other math topics
10 guides
keyboard_arrow_down
check_circle
Mark as learned
thumb_up
0
thumb_down
0
chat_bubble_outline
0
Comment
auto_stories Bi-column layout
settings

# Properties of complex conjugate with proofs

schedule Aug 12, 2023
Last updated
local_offer
Tags
mode_heat
Master the mathematics behind data science with 100+ top-tier guides
Start your free 7-days trial now!
Definition.

# Complex conjugates

The complex numbers $a+bi$ and $a-bi$ are known as complex conjugates. If we let $z=a+bi$, then the complex conjugate of $z$ is denoted as $\bar{z}=a-bi$ or $\overline{a+bi}= a-bi$.

Example.

## Finding the complex conjugate (1)

Find the complex conjugate of $z=3+2i$.

Solution. The complex conjugate of $z=3+2i$ is:

$$z=3-2i$$

This completes the proof.

Example.

## Finding the complex conjugate (2)

Find the complex conjugate of $5+8i$.

Solution. The complex conjugate of $5+8i$ is:

$$\overline{5+8i}=5-8i$$

# Properties of complex conjugate

Theorem.

## Sum of a complex number and its complex conjugate

If $z=a+bi$, then $z+\overline{z}$ is:

$$z+\overline{z}=2a$$

This means that $z+\overline{z}$ is a real number.

Proof. If $z=a+bi$, then $z+\overline{z}$ is:

\begin{align*} z+\overline{z} &=a+bi+(a-bi)\\ &=2a \end{align*}

This completes the proof.

Theorem.

## Product of a complex number and its complex conjugate

If $z=a+bi$, then the product $z\overline{z}$ is:

$$z\bar{z}=a^2+b^2$$

This means that the product of a complex number and its complex conjugate results in a purely real numberlink.

This is the trick we used to get rid of the imaginary part in the denominator when dividing two complex numbers!

Proof. The product $z\bar{z}$ is:

\begin{align*} z\bar{z}&= (a+bi)(a-bi)\\&= a^2-abi+abi-b^2i^2\\ &=a^2+b^2 \end{align*}

Here, for the second equality, we used the fact that $i^2=-1$. This completes the proof.

Example.

## Multiplying a complex number and its complex conjugate

Let $z=3+4i$. Compute $z\bar{z}$.

Solution. By theoremlink, we have that:

\begin{align*} z\bar{z}&=3^2+4^2\\ &=25 \end{align*}
Theorem.

## Complex conjugate of a sum of two complex numbers

If $z_1$ and $z_2$ are complex numbers, then the complex conjugate of $z_1+z_2$ is:

$$\overline{z_1+z_2}= \overline{z_1}+\overline{z_2}$$

Proof. Let $z_1$ and $z_2$ be defined as follows:

\begin{align*} z_1&=a+bi\\ z_2&=c+di \end{align*}

Now, $\overline{z_1+z_2}$ is:

\begin{align*} \overline{z_1+z_2}&= \overline{(a+bi)+(c+di)}\\ &=\overline{(a+c)+(b+d)i}\\ &=(a+c)-(b+d)i\\ &=(a-bi)+(c-di)\\ &=\overline{z_1}+\overline{z_2} \end{align*}

This completes the proof.

Theorem.

## Complex conjugate of a product of two complex numbers

If $z_1$ and $z_2$ are complex numbers, then the complex conjugate of $z_1z_2$ is:

$$\overline{z_1z_2}= \overline{z_1}\cdot\overline{z_2}$$

Proof. Let $z_1$ and $z_2$ be defined as follows:

\begin{align*} z_1&=a+bi\\ z_2&=c+di \end{align*}

Now, $\overline{z_1z_2}$ is:

\begin{align*} \overline{z_1z_2}&= \overline{(a+bi)(c+di)}\\ &=\overline{ac+adi+bci+bdi^2}\\ &=\overline{(ac-bd)+(ad+bc)i}\\ &=ac-bd-(ad+bc)i\\ &=ac-adi-bci+bdi^2\\ &=(a-bi)(c-di)\\ &=\overline{z_1}\cdot\overline{z_2} \end{align*}

This completes the proof.

Theorem.

## Complex conjugate of a product between a real and complex number

If $k$ is a real number and $z$ is a complex number, then:

$$\overline{kz}= k\overline{z}$$

Proof. If $z=a+bi$ and $k\in\mathbb{R}$, then $\overline{kz}$ is:

\begin{align*} \overline{kz}&=\overline{k(a+bi)}\\ &=\overline{ak+bki}\\ &=ak-bki\\ &=k(a-bi)\\ &=k\overline{z} \end{align*}

This completes the proof.

Theorem.

## Complex conjugate of a complex conjugate is equal to the complex number itself

If $\bar{z}$ is the complex conjugate of $z$, then the complex conjugate of $\bar{z}$ is equal to $z$, that is:

$$\overline{\bar{z}}=z$$

Proof. Let $z=a+bi$. The complex conjugate of $z$ is:

$$\bar{z}=a-bi$$

The complex conjugate of $\bar{z}$ is:

\begin{align*} \overline{\bar{z}} &=a+bi\\ &=z \end{align*}

This completes the proof.

Theorem.

## Complex conjugate of the sum of products of real and complex numbers

If $k_1$, $k_2$, $\cdots$, $k_n$ are real numbers and $z_1$, $z_2$, $\cdots$, $z_n$ are complex numbers, then:

$$\overline{k_1z_1+k_2z_2+\cdots+k_nz_n}= k_1\overline{z_1}+k_2\overline{z_2}+ \cdots+k_n\overline{z_n}$$

Proof. Firstly, we repeatedly apply propertylink to get:

\begin{align*} \overline{k_1z_1+k_2z_2+k_3z_3+\cdots+k_nz_n}&= \overline{k_1z_1+(k_2z_2+k_3z_3+\cdots+k_nz_n)}\\&= \overline{k_1z_1}+\overline{(k_2z_2+k_3z_3+\cdots+k_nz_n)}\\ &=\overline{k_1z_1}+\overline{(k_2z_2)+(k_3z_3+\cdots+k_nz_n)}\\ &=\overline{k_1z_1}+\overline{k_2z_2}+\overline{k_3z_3+\cdots+k_nz_n}\\ &=\overline{k_1z_1}+\overline{k_2z_2}+ \overline{k_3z_3}\cdots+\overline{k_nz_n}\\ \end{align*}

For each complex conjugate term, we apply propertylink to get:

\begin{align*} \overline{k_1z_1+k_2z_2+k_3z_3+\cdots+k_nz_n} &=\overline{k_1z_1}+\overline{k_2z_2}+\overline{k_3z_3}+\cdots+\overline{k_nz_n}\\ &=k_1\overline{z_1}+k_2\overline{z_2}+k_3\overline{z_3}+ \cdots+k_n\overline{z_n}\\ \end{align*}

This completes the proof.

Theorem.

## If a complex number and its conjugate are equal, then the complex number is real

Let $z$ be a complex number and $\overline{z}$ be its complex conjugate. If $z=\overline{z}$, then $z$ is a real number.

Proof. Let $z=a+bi$. The complex conjugate of $z$ is $\overline{z}=a-bi$. Equating the two gives:

\begin{align*} a+bi&=a-bi\\ 2bi&=0\\ b&=0 \end{align*}

This means that $z=a$ and $\overline{z}=a$, that is, they do not have an imaginary part. A complex number without the imaginary part is a real number, and so $z$ and $\overline{z}$ are both real numbers. This completes the proof.

Theorem.

## Relationship between complex conjugate and modulus

If $z$ is a complex number, then:

$$z\overline{z}=\vert{z}\vert^2$$

Where $\vert{z}\vert$ is the moduluslink of $z$.

Proof. Let $z=a+bi$. By propertylink, $z\overline{z}$ is:

\label{eq:t21aHvVZhYuvJXxqe3k} \begin{aligned}[b] z\overline{z}&=a^2+b^2 \end{aligned}

Now, by definitionlink, the modulus of $z$ is:

$$\vert{z}\vert=\sqrt{a^2+b^2}$$

Taking the square of both sides gives:

$$$$\label{eq:DbzmTbhuBM2pcXJDD8f} \vert{z}\vert^2=a^2+b^2$$$$

Equating \eqref{eq:t21aHvVZhYuvJXxqe3k} and \eqref{eq:DbzmTbhuBM2pcXJDD8f} gives:

$$z\overline{z}=\vert{z}\vert^2$$

This completes the proof.

Edited by 0 others
thumb_up
thumb_down
Comment
Citation
Ask a question or leave a feedback...
thumb_up
0
thumb_down
0
chat_bubble_outline
0
settings
Enjoy our search
Hit / to insta-search docs and recipes!