If we multiply a complex number z=${\mathbf{re}}^{\mathbf{i\varphi }}$by ${\mathbf{e}}^{\mathbf{i\varphi }}$, we get ${\mathbf{e}}^{\mathbf{i\theta }}\mathbf{Z}\mathbf{=}{\mathbf{re}}^{\mathbf{i}\left(\varphi +\theta \right)}$, that is, a complex number with the same but with its angle increased by $\mathbf{\theta }$ . We can say that the vector from the origin to the point $\mathbf{z}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{iy}$has been rotated by angle $\mathbf{\theta }$as in Figure 7.4to become the vectorsRfrom the origin to the point $\mathbf{z}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{iy}$. Then we can write$\mathbf{X}\mathbf{+}\mathbf{iY}\mathbf{=}{\mathbf{e}}^{\mathbf{i\theta }}\mathbf{z}\mathbf{=}{\mathbf{e}}^{\mathbf{i\theta }}\left(x+\mathrm{iy}\right)$ . Take real and imaginary parts of this equation to obtain equations$\left(7.12\right)$ .

The relation between the exponential function $e^{i\theta }$and the trigonometric functions is

role="math" localid="1658997426767" ${\mathrm{e}}^{\mathrm{i\theta }}=\mathrm{cos\theta }+\mathrm{isin\theta }$.

The complex number given is $z=r{e}^{i\theta }$and the vector r from the origin to the point$\mathrm{z}=\mathrm{x}+\mathrm{iy}$

The complex number $z=r{e}^{i\theta }$in the polar form is given by $\mathrm{z}=\mathrm{r}\left(\mathrm{cos\varphi }+\mathrm{isin\varphi }\right)$.

We multiply the complex number $\mathrm{z}={\mathrm{re}}^{\mathrm{i\varphi }}$by $e^{i\theta }$,and we get

$$\begin{gathered} {\mathrm{e}}^{\mathrm{i\theta }}\mathrm{Z}={\mathrm{e}}^{\mathrm{i\theta }}{\mathrm{e}}^{\mathrm{i\varphi }} \\ ={\mathrm{re}}^{\mathrm{i}\left(\mathrm{\varphi }+\mathrm{\theta }\right)} \end{gathered}$$

The vector $\overrightarrow{r}$which from the origin to the point z=x+iy has been rotated by an angle $\theta$to become the vector $\overrightarrow{R}$from the origin to the complex number $Z=X+iY$.

We can write $\mathrm{X}+\mathrm{iY}={\mathrm{e}}^{\mathrm{i\theta }}\mathrm{z}={\mathrm{e}}^{\mathrm{i\theta }}\left(\mathrm{x}+\mathrm{iy}\right)$implies that

$$\begin{gathered} Z=X+iY \\ =e^{i\theta }z \\ =e^{i\theta }\left(x+iy\right) \end{gathered}$$

As the relation between the exponential function $e^{i\theta }$and the trigonometric functions implies that $e^{i\theta }=\cos \theta +i\sin \theta$.

$$\begin{gathered} Z=X+iY \\ =e^{i\theta }z \\ =e^{i\theta }\left(x+iy\right) \\ =\left(\cos \theta +i\sin \theta \right)\left(x+iy\right) \end{gathered}$$

To further solve,

$$\begin{gathered} \mathrm{Z}=\left(\mathrm{x}\right)\mathrm{cos\theta }+\mathrm{i}\left(\mathrm{x}\right)\mathrm{sin\theta }+\mathrm{i}\left(\mathrm{y}\right)\mathrm{cos\theta }+\left(\mathrm{i}\right)\left(\mathrm{i}\right)\left(\mathrm{y}\right)\mathrm{sin\theta } \\ =\mathrm{xcos\theta }+\mathrm{ixsin\theta }+\mathrm{iycos\theta }+{\mathrm{i}}^2\mathrm{ysin\theta } \\ =\left(\mathrm{xcos\theta }-\mathrm{ysin\theta }\right)+\mathrm{i}\left(\mathrm{xsin\theta }+\mathrm{ycos\theta }\right) \end{gathered}$$

Therefore, we deduce the following set of the equation,

$$\begin{gathered} \mathrm{X}=\mathrm{xcos\theta }-\mathrm{ysin\theta } \\ \mathrm{Y}=\mathrm{xsin\theta }+\mathrm{ycos\theta } \end{gathered}$$

The vector $\overrightarrow{r}$which from the origin to the point z=x+iy has been rotated by an angle $\theta$to become the vector $\overrightarrow{R}$from the origin to the point $Z=X+iY$ written in the matrix form $\left(\begin{array}{c}X\\ Y\end{array}\right)=\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)$, vector rotated which relates the components of$\overrightarrow{r}and\overrightarrow{R}$.