Question

Prove that the conjugate of the quotient of two complex numbers is the quotient of the conjugates. Also prove the corresponding statements for difference and product. Hint: It is easier to prove the statements about product and quotient using the polar coordinate form; for the difference, $\mathit{r}{\mathit{e}}^{\mathbf{i}\mathbf{\theta }}$it is easier to use the rectangularform .$\mathit{x}\mathbf{+}\mathit{i}\mathit{y}$

Short answer

It is proved that the conjugate of the quotient, product anddifferences of two complex numbers is the quotient, product anddifferencesof the conjugates respectively.

Step-by-step solution

Define the complex Conjugate  

For any two complex numbers, let say $z_1\text{and}{z}_2$, the sum of their conjugate is given by: .${\overline{\mathbf{z}}}_{\mathbf{1}}\mathbf{+}{\overline{\mathbf{z}}}_{\mathbf{2}}\mathbf{=}\overline{\mathbf{(}{\mathbf{z}}_{\mathbf{1}}\mathbf{+}{\mathbf{z}}_{\mathbf{2}}\mathbf{)}}$

Step 2:Derive the proof

Let us assume two complex numbers as$z_1\text{and}{z}_2$: where,

$\begin{array}{l}{z}_1=r_1{e}^{i{\theta }_1}\Rightarrow {\overline{z}}_1=r_1{e}^{-i{\theta }_1}\\ z_2=r_2{e}^{i{\theta }_2}\Rightarrow {\overline{z}}_2=r_2{e}^{-i{\theta }_2}\end{array}$

Then, their ratio will be:

$\begin{array}{c}\frac{z_1}{z_2}=\frac{r_1{e}^{i{\theta }_1}}{r_2{e}^{i{\theta }_2}}\\ =\frac{r_1}{r_2}{e}^{i({\theta }_1-{\theta }_2)}\end{array}$

And the conjugate of this ratio will be:

$\overline{\left(\frac{z_1}{z_2}\right)}=\frac{r_1}{r_2}{e}^{-i({\theta }_1-{\theta }_2)}$

Also, the ratio of their conjugate will be

$\begin{array}{c}\frac{{\overline{z}}_1}{{\overline{z}}_2}=\frac{r_1{e}^{-i{\theta }_1}}{r_2{e}^{-i{\theta }_2}}\\ =\frac{r_1}{r_2}{e}^{-i({\theta }_1-{\theta }_2)}\end{array}$

Clearly, we see: $\overline{\left(\frac{z_1}{z_2}\right)}=\frac{{\overline{z}}_1}{{\overline{z}}_2}$

Hence proved, the conjugate of the quotient of two complex numbers is the quotient of the conjugates.

Again, we have:

$\begin{array}{l}{z}_1=r_1{e}^{i{\theta }_1}\Rightarrow {\overline{z}}_1=r_1{e}^{-i{\theta }_1}\\ z_2=r_2{e}^{i{\theta }_2}\Rightarrow {\overline{z}}_2=r_2{e}^{-i{\theta }_2}\end{array}$

Then, their product will be:

$\begin{array}{c}{z}_1{z}_2=\left(r_1{e}^{i{\theta }_1}\right)\left(r_2{e}^{i{\theta }_2}\right)\\ =r_1{r}_2{e}^{i({\theta }_1+{\theta }_2)}\end{array}$

And the conjugate of this obtained product will be:

$\begin{array}{c}\overline{z_1{z}_2}=\overline{r_1{r}_2{e}^{i({\theta }_1+{\theta }_2)}}\\ =r_1{r}_2{e}^{-i({\theta }_1+{\theta }_2)}\end{array}$

Also, the product of their conjugate will be

$\begin{array}{c}{\overline{z}}_1{\overline{z}}_2=\left(r_1{e}^{-i{\theta }_1}\right)\left(r_2{e}^{-i{\theta }_2}\right)\\ =r_1{r}_2{e}^{-i({\theta }_1+{\theta }_2)}\end{array}$

Clearly, we see: $\overline{z_1{z}_2}={\overline{z}}_1{\overline{z}}_2$

Hence proved, the conjugate of the product of two complex numbers is the product of the conjugates.

Now, for the difference, let us use rectangular form as:

$\begin{array}{l}{z}_1=x_1+i{y}_1\Rightarrow {\overline{z}}_1=x_1-i{y}_1\\ z_2=x_2+i{y}_2\Rightarrow {\overline{z}}_2=x_2-i{y}_2\end{array}$

Then, their differences will be:

$\begin{array}{c}{z}_1-z_2=(x_1+i{y}_1)-(x_2+i{y}_2)\\ =(x_1-x_2)+i(y_1-y_2)\end{array}$

And the conjugate of this obtained expression will be:

$\begin{array}{c}\overline{z_1-z_2}=\overline{(x_1-x_2)+i(y_1-y_2)}\\ =(x_1-x_2)-i(y_1-y_2)\end{array}$

Also, the differences of their conjugate will be

$\begin{array}{c}{\overline{z}}_1-{\overline{z}}_2=(x_1-i{y}_1)-(x_2-i{y}_2)\\ =(x_1-x_2)-i(y_1-y_2)\end{array}$

Clearly, we see: $\overline{z_1-z_2}={\overline{z}}_1-{\overline{z}}_2$

Hence proved, the conjugate of the differences of two complex numbers is the differences of the conjugates.