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.

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Example.

Finding the complex conjugate (2)

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

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

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

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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.

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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.

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Example.

Multiplying a complex number and its complex conjugate

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

Solution. By theoremlink, we have that:

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

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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.

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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.

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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.

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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.

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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.

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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.

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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:

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

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

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

Taking the square of both sides gives:

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

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

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

This completes the proof.

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Published by Isshin Inada

Edited by 0 others

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