Question

There is a one-to-one correspondence between two-dimensional vectors and complex numbers. Show that the real and imaginary parts of the product ${\mathit{z}}_{\mathbf{1}}{\mathit{z}}_{\mathbf{2}}^{\mathbf{*}}$(the star denotes complex conjugate) are respectively the scalar product and$\mathbf{\pm }$the magnitude of the vector product of the vectors corresponding to${\mathbf{z}}_{\mathbf{1}}$and${\mathbf{z}}_{\mathbf{2}}$.

Short answer

The real and imaginary parts of product${\mathrm{z}}_1{\mathrm{z}}_2$* of two complex numbers and complex conjugate $\left({\mathrm{z}}_2\right)*$of ${\mathrm{z}}_2$are respectively the scalar product and $\pm$the magnitude of vector product of the vectors corresponding to complex numbers $z_1$and $z_2$.

Step-by-step solution

Given information.

There is a one-to-one correspondence between two-dimensional vectors and complex numbers ${\mathrm{z}}_1$and${\mathrm{z}}_2$

Complex conjugate of a complex number and product of two complex numbers.

The complex conjugate of a complex number z = x + iyis a complex number $\mathbf{\otimes }$.

The product of two complex numbers a + iband c + idis given by (ac-bd) +i (ad + bc).

Find the vector corresponding to the complex number.

Let ${\mathrm{z}}_1={\mathrm{x}}_1+{\mathrm{iy}}_1$and ${\mathrm{z}}_2={\mathrm{x}}_2+{\mathrm{iy}}_2$are two complex numbers geometrically shown in figure here.

The conjugate of${\mathrm{z}}_2={\mathrm{x}}_2+{\mathrm{iy}}_2$is${\mathrm{z}}_2*={\mathrm{x}}_2-{\mathrm{iy}}_2$. Then the product$z_1{z}_2*$is$$\begin{gathered} {\mathrm{z}}_1{\mathrm{z}}_2*=\left({\mathrm{x}}_1+{\mathrm{iy}}_1\right)\left({\mathrm{x}}_2+{\mathrm{iy}}_2\right) \\ =\left({\mathrm{x}}_1{\mathrm{x}}_2+{\mathrm{y}}_1{\mathrm{y}}_2\right)+\mathrm{i}\left(-{\mathrm{x}}_1{\mathrm{y}}_2+{\mathrm{x}}_2{\mathrm{y}}_1\right) \end{gathered}$$

Thus,

$$\begin{gathered} \mathrm{Re}\left({\mathrm{z}}_1{\mathrm{z}}_2^{*}\right)=\left({\mathrm{x}}_1{\mathrm{x}}_2+{\mathrm{y}}_1+{\mathrm{y}}_2\right)...\left(1\right) \\ \mathrm{Im}\left({\mathrm{z}}_1{\mathrm{z}}_2^{*}\right)=\left(-{\mathrm{x}}_1{\mathrm{y}}_2+{\mathrm{x}}_2+{\mathrm{y}}_1\right)...\left(2\right) \end{gathered}$$

The points $\mathrm{P}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$and $Q\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$are shown in figure in x - y plane. Then the vector corresponding to complex number ${\mathrm{z}}_1={\mathrm{x}}_1+{\mathrm{iy}}_1\mathrm{is}\overrightarrow{\mathrm{OP}}={\mathrm{x}}_1{\hat{\mathrm{e}}}_{\mathrm{x}}+{\mathrm{y}}_1{\hat{\mathrm{e}}}_{\mathrm{y}}$and the vector corresponding to complex number ${\mathrm{z}}_2={\mathrm{x}}_2+{\mathrm{iy}}_2\mathrm{is}\overrightarrow{OQ}={\mathrm{x}}_2{\hat{\mathrm{e}}}_{\mathrm{x}}+{\mathrm{y}}_2{\hat{\mathrm{e}}}_{\mathrm{y}}$, where ${\hat{\mathrm{e}}}_{\mathrm{x}}$and ${\hat{\mathrm{e}}}_{\mathrm{y}}$are unit vectors along -axis and -axis respectively.

Find the scalar and vector product.

Now, the scalar products of vectors$\overline{\mathrm{O}}\overrightarrow{\mathrm{P}}$and$\overline{\mathrm{O}}\overrightarrow{\mathrm{Q}}$is

$$\begin{gathered} \overline{\mathrm{O}}\overrightarrow{\mathrm{P}}.\overline{\mathrm{O}}\overrightarrow{\mathrm{Q}}=\left(x_1{\hat{e}}_x+y_1{\hat{e}}_y\right)\left(x_2{\hat{e}}_x+y_2{\hat{e}}_y\right) \\ =x_1{x}_2+y_1{y}_2 \\ =Re\left(z_1{z}_2^{*}\right)...\left(3\right) \\ \overline{\mathrm{O}}\overrightarrow{\mathrm{P}}.\overline{\mathrm{O}}\overrightarrow{\mathrm{Q}}=\left(x_2{\hat{e}}_x+y_2{\hat{e}}_y\right)\left(x_1{\hat{e}}_x+y_1{\hat{e}}_y\right) \\ =x_2{x}_1+y_2{y}_1 \\ =Re\left(z_1{z}_2^{*}\right)...\left(4\right) \end{gathered}$$

Now, the vector products of vectors $\overline{\mathrm{O}}\overrightarrow{\mathrm{P}}$and $\overline{\mathrm{O}}\overrightarrow{\mathrm{Q}}$is

$$\begin{gathered} \overline{\mathrm{O}}\overrightarrow{\mathrm{P}}.\overline{\mathrm{O}}\overrightarrow{\mathrm{Q}}=\left(x_1{\hat{e}}_x+y_1{\hat{e}}_y\right)\times \left(x_2{\hat{e}}_x+y_2{\hat{e}}_y\right) \\ =\left|\begin{array}{cc}{x}_1& y_1\\ x_2& y_2\end{array}\right| \\ =\left(x_1{y}_2-x_2{y}_1\right) \\ =-\left(-x_1{y}_2+x_2{y}_1\right) \end{gathered}$$

Hence the product is _{$=-Im\left(z_1{z}_2^{*}\right)...\left(5\right)$}

_{$$\begin{gathered} \overline{\mathrm{O}}\overrightarrow{\mathrm{P}}\times \overline{\mathrm{O}}\overrightarrow{\mathrm{Q}}=\left(x_2{\hat{e}}_x+y_2{\hat{e}}_y\right)\times \left(x_1{\hat{e}}_x+y_1{\hat{e}}_y\right) \\ =\left|\begin{array}{cc}{x}_2& y_2\\ x_1& y_1\end{array}\right| \\ =\left(x_2{y}_1-x_1{y}_2\right) \\ =-\left(-x_1{y}_2+x_2{y}_1\right) \end{gathered}$$}

Hence the product is $=\mathrm{Im}\left({\mathrm{z}}_1{\mathrm{z}}_2^{*}\right)...\left(6\right)$

From equations (1) to (6), one can see that the real and imaginary parts of product$z_1{z}_2*$ of two complex numbers$z_1$ and complex conjugate$\left(z_2\right)$* of$z_2$ are respectively the scalar product and $\pm$the magnitude of vector product of the vectors corresponding to complex numbers$z_1$ and $z_2$.

Hence, proved.