91Ó°ÊÓ

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

role="math" localid="1658997426767" ${\mathrm{e}}^{\mathrm{¾\pm θ}}=\mathrm{³¦´Ç²õθ}+\mathrm{¾\pm ²õ¾\pm ²Ôθ}$.

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

The complex number $z=r{e}^{i\mathrm{θ}}$in the polar form is given by $\mathrm{z}=\mathrm{r}\left(\mathrm{³¦´Ç²õÏ•}+\mathrm{¾\pm ²õ¾\pm ²ÔÏ•}\right)$.

We multiply the complex number $\mathrm{z}={\mathrm{re}}^{\mathrm{¾\pm Ï•}}$by $e^{i\mathrm{θ}}$,and we get

$$\begin{gathered} {\mathrm{e}}^{\mathrm{¾\pm θ}}\mathrm{Z}={\mathrm{e}}^{\mathrm{¾\pm θ}}{\mathrm{e}}^{\mathrm{¾\pm Ï•}} \\ ={\mathrm{re}}^{\mathrm{i}\left(\mathrm{Ï•}+\mathrm{θ}\right)} \end{gathered}$$

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

We can write $\mathrm{X}+\mathrm{iY}={\mathrm{e}}^{\mathrm{¾\pm θ}}\mathrm{z}={\mathrm{e}}^{\mathrm{¾\pm θ}}\left(\mathrm{x}+\mathrm{iy}\right)$implies that

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

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

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

To further solve,

$$\begin{gathered} \mathrm{Z}=\left(\mathrm{x}\right)\mathrm{³¦´Ç²õθ}+\mathrm{i}\left(\mathrm{x}\right)\mathrm{²õ¾\pm ²Ôθ}+\mathrm{i}\left(\mathrm{y}\right)\mathrm{³¦´Ç²õθ}+\left(\mathrm{i}\right)\left(\mathrm{i}\right)\left(\mathrm{y}\right)\mathrm{²õ¾\pm ²Ôθ} \\ =\mathrm{x³¦´Ç²õθ}+\mathrm{ix²õ¾\pm ²Ôθ}+\mathrm{iy³¦´Ç²õθ}+{\mathrm{i}}^2\mathrm{y²õ¾\pm ²Ôθ} \\ =\left(\mathrm{x³¦´Ç²õθ}-\mathrm{y²õ¾\pm ²Ôθ}\right)+\mathrm{i}\left(\mathrm{x²õ¾\pm ²Ôθ}+\mathrm{y³¦´Ç²õθ}\right) \end{gathered}$$

Therefore, we deduce the following set of the equation,

$$\begin{gathered} \mathrm{X}=\mathrm{x³¦´Ç²õθ}-\mathrm{y²õ¾\pm ²Ôθ} \\ \mathrm{Y}=\mathrm{x²õ¾\pm ²Ôθ}+\mathrm{y³¦´Ç²õθ} \end{gathered}$$

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