Q25P - Mathematical Methods in Physical Sciences

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 rectangular form .$\mathit{x}\mathbf{+}\mathit{i}\mathit{y}$

Step-by-Step Solution

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 1: 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.