Trigonometric and exponential forms of a complex number.

Literature: Collection of problems in mathematics. Part 1. Ed. by A. V. Efimov, B. P. Demidovich.

Trigonometric form of a complex number:

For any complex number $z=x+iy$, the equality holds: $$z=|z|(\cos\varphi+i\sin\varphi).\qquad\qquad\qquad (1)$$ Here $|z|=\sqrt{x^2+y^2}$, and $\varphi$ satisfies the conditions: $$\cos\varphi=\frac{x}{\sqrt{x^2+y^2}},\qquad \sin\varphi=\frac{y}{\sqrt{x^2+y^2}},\qquad \varphi\in[0, 2\pi).$$

The equality (1) is called the trigonometric form of the complex number $z$.

Examples:

Represent the following complex numbers in trigonometric form and plot them as points on the complex plane:

1.435. $-i$

Solution:

Let $z=x+iy=-i$, which means $x=0$ and $y=-1$. Then $$|z|=\sqrt{x^2+y^2}=\sqrt 1=1.$$

$$\cos\varphi=\frac{0}{1}=0,\qquad \sin\varphi=\frac{-1}{1}=-1\Rightarrow \varphi=\frac{3\pi}{2}.$$

Thus, $z=\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2}$.

Answer: $\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2}$.

1.438. $\frac{1-i}{1+i}$.

Solution.

Let's express the number $z=\frac{1-i}{1+i}$ in algebraic form:

$$\frac{1-i}{1+i}=\frac{(1-i)(1-i)}{(1+i)(1-i)}=\frac{1-2i+i^2}{1-i^2}=\frac{1-2i-1}{1+1}=\frac{-2i}{2}=-i.$$

The trigonometric form of the number $-i$ was found in the previous example (1.435):

$z=-i=\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2}$.

Answer: $\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2}$.

1.441. $1+\cos\frac{\pi}{7}+i\sin\frac{\pi}{7}$.

Solution.

Let $z=x+iy=1+\cos\frac{\pi}{7}+i\sin{\pi}{7}$, so $x=1+\cos\frac{\pi}{7},,, y=\sin{\pi}{7}$. Then$$|z|=\sqrt{x^2+y^2}=\sqrt {\left(1+\cos\frac{\pi}{7}\right)^2+\sin^2\frac{\pi}{7}}=$$ $$=\sqrt{1+2\cos\frac{\pi}{7}+\cos^2\frac{\pi}{7}+\sin^2\frac{\pi}{7}}=\sqrt{2+2\cos\frac{\pi}{7}}=$$ $$=\sqrt{4\cos^2\frac{\pi}{14}}=2\cos\frac{\pi}{14}.$$

$$\cos\varphi=\frac{x}{|z|}=\frac{1+\cos\frac{\pi}{7}}{2\cos\frac{\pi}{14}}=\frac{2\cos^2\frac{\pi}{14}}{2\cos\frac{\pi}{14}}=\cos\frac{\pi}{14}.$$

$$\sin\varphi=\frac{y}{|z|}=\frac{sin\frac{\pi}{7}}{2\cos\frac{\pi}{14}}=\frac{2\cos\frac{\pi}{14}\sin\frac{\pi}{14}}{2\cos\frac{\pi}{14}}=\sin\frac{\pi}{14}.$$

Thus, $\varphi=\frac{\pi}{14}$.

From here, we find the exponential form of the complex number $z=x+iy=1+\cos\frac{\pi}{7}+i\sin{\pi}{7}:$

$$z=2\cos\frac{\pi}{14}\left(\cos\frac{\pi}{14}+i\sin\frac{\pi}{14}\right).$$

Answer: $2\cos\frac{\pi}{14}\left(\cos\frac{\pi}{14}+i\sin\frac{\pi}{14}\right).$

Exponential form of a complex number:

The symbol $e^{i\varphi}$ denotes the complex number $\cos\varphi+i\sin\varphi$. Using this notation, any complex number $z=|z|(\cos\varphi+i\sin\varphi)$ can be represented in exponential form as $$z=|z|e^{i\varphi}.$$

Examples.

Represent the following complex numbers in exponential form:

1.475. $\frac{7+24i}{5}$.

Solution.

Let's express the number $z=\frac{7+24i}{5}$ in algebraic form:

$$z=x+iy=\frac{7+24i}{5}=\frac{7}{5}+\frac{24}{5}i.$$

$$|z|=\sqrt{\left(\frac{7}{5}\right)^2+\left(\frac{24}{5}\right)^2}=\sqrt{\frac{49+576}{25}}=\sqrt{\frac{625}{25}}=\sqrt{25}=5.$$

$$tg\varphi=\frac{y}{x}=\frac{\frac{24}{5}}{\frac{7}{5}}=\frac{24}{7}.$$ Поскольку число $z$ принадлежит первой четверти, то $\varphi=arctg\frac{24}{7}.$

Thus, $z=5e^{i \arctan\frac{24}{7}}$.

Answer: $z=5e^{i \arctan\frac{24}{7}}$.

1.479. $\sin\alpha-i\cos\alpha$.

Solution.

$$z=x+iy=\sin\alpha-i\cos\alpha\Rightarrow \,\,x=\sin\alpha,\,\,y=-cos\alpha.$$

$$|z|=\sqrt{x^2+y^2}=\sqrt{\sin^2\alpha+\cos^2\alpha}=1.$$

$$tg\varphi=\frac{y}{x}=\frac{-\cos\alpha}{\sin\alpha}=-ctg\alpha=tg(\alpha+\frac{\pi}{2})=tg(\alpha+\frac{3\pi}{2}).$$

Additionally, the following conditions must be satisfied:

$$\cos\varphi=\frac{x}{|z|}=\sin\alpha;\qquad \sin\varphi=\frac{y}{|z|}=\cos\alpha.$$

From here, we find

$$\varphi=\alpha+\frac{3\pi}{2}.$$

Thus, $$z=\sin\alpha-i\cos\alpha=e^{i\left(\alpha+\frac{3\pi}{2}\right)}.$$

Answer: $e^{i\left(\alpha+\frac{3\pi}{2}\right)}.$

1.482 (a). Express the numbers $z_1$ and $z_2$ in exponential form and perform the specified operations on them:

$z_1z_2;$ $\frac{z^2_1}{z_2},$ if $z_1=2\sqrt 3-2i,$ $z_2=3-3\sqrt 3i.$

Solution.

Let's express the numbers $z_1$ and $z_2$ in exponential form:

$$|z_1|=\sqrt{x^2+y^2}=\sqrt{(2\sqrt 3)^2+(-2)^2}=\sqrt{16}=4.$$

$$tg\varphi=\frac{y}{x}=\frac{-2}{2\sqrt 3}=-\frac{1}{\sqrt 3}.$$

Since the number $z_1$ belongs to the fourth quadrant, then $\varphi_1=\arctan{\left(-\frac{1}{\sqrt 3}\right)}=-\frac{\pi}{6}$.

From here, $$z_1=4e^{-i\frac{\pi}{6}}.$$

$$|z_2|=\sqrt{x^2+y^2}=\sqrt{3^2+(-3\sqrt 3)^2}=\sqrt{36}=6.$$

$$tg\varphi=\frac{y}{x}=\frac{-3\sqrt 3}{3}=-\sqrt 3.$$

Since the number $z_2$ belongs to the fourth quadrant, then $\varphi_2=\arctan{\sqrt{3}}=\frac{\pi}{3}$.

From here, $$z_2=6e^{-i\frac{\pi}{3}}.$$

Next, we find $z_1z_2$ and $\frac{z_1^2}{z_2}$:

$$z_1z_2=4e^{-i\frac{\pi}{6}}6e^{-\frac{\pi}{3}}=24e^{i\left(\frac{-\pi}{6}-\frac{\pi}{3}\right)}=24e^{-i\frac{\pi}{2}}=$$

$$=24\left(\cos\left(-\frac{\pi}{2}\right)+i\sin\left(-\frac{\pi}{2}\right)\right)=24(0-1)=-24.$$

$$\frac{z^2_1}{z_2}=\frac{(4e^{-i\frac{\pi}{6}})^2}{6e^{-\frac{\pi}{3}}}=\frac{16}{6}e^{i\left(\frac{-2\pi}{6}+\frac{\pi}{3}\right)}=\frac{8}{3}e^{i\cdot 0}=\frac{8}{3}.$$

Answer: $-24, \frac{8}{3}.$

Homework.

Represent the following complex numbers in trigonometric form and plot them as points on the complex plane:

1.436. $1-i\sqrt{3}$.

Answer: $2\left(\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}\right)$.

1.437. $-\frac{1}{2}+i\frac{\sqrt{3}}{2}$.

Answer: $\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}$.

1.440. $\sin\frac{\pi}{3}+i\cos\frac{\pi}{3}$.

Answer: $\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}$.

Represent the following complex numbers in exponential form:

1.476. $5-12i$.

Answer: $13e^{-i \arctan\left(-\frac{12}{5}\right)}$.