Trigonometric and exponential forms of a complex number.

Trigonometric form of a complex number:

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

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. $-i$

Solution:

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

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

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

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

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

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

3. $1+\cos \frac{\pi }{7}+i\sin \frac{\pi }{7}$.

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$$|z|=\sqrt{x^2+y^2}=\sqrt{{(1+\cos \frac{\pi }{7})}^2+{\sin }^2\frac{\pi }{7}}=$$ $$=\sqrt{1+2\cos \frac{\pi }{7}+{\cos }^2\frac{\pi }{7}+{\sin }^2\frac{\pi }{7}}=\sqrt{2+2\cos \frac{\pi }{7}}=$$ $$=\sqrt{4{\cos }^2\frac{\pi }{14}}=2\cos \frac{\pi }{14}.$$

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

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

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

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

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

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

Exponential form of a complex number:

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

Examples.

Represent the following complex numbers in exponential form:

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

Solution.

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

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

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

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

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

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

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

Solution.

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

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

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

Additionally, the following conditions must be satisfied:

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

From here, we find

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

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

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

5. Express the numbers $z_1$ and $z_2$ in exponential form and perform the specified operations on them:

$z_1{z}_2;$ $\frac{z_1^2}{z_2},$ if $z_1=2\sqrt{3}-2i,$ $z_2=3-3\sqrt{3}i.$

Solution.

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

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

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

Since the number $z_1$ belongs to the fourth quadrant, then ${\phi }_1=\arctan (-\frac{1}{\sqrt{3}})=-\frac{\pi }{6}$.

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

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

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

Since the number $z_2$ belongs to the fourth quadrant, then ${\phi }_2=\arctan \sqrt{3}=\frac{\pi }{3}$.

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

Next, we find $z_1{z}_2$ and $\frac{z_1^2}{z_2}$:

$$z_1{z}_2=4e^{-i\frac{\pi }{6}}6e^{-\frac{\pi }{3}}=24e^{i(\frac{-\pi }{6}-\frac{\pi }{3})}=24e^{-i\frac{\pi }{2}}=$$

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

$$\frac{z_1^2}{z_2}=\frac{(4e^{-i\frac{\pi }{6}}{)}^2}{6e^{-\frac{\pi }{3}}}=\frac{16}{6}{e}^{i(\frac{-2\pi }{6}+\frac{\pi }{3})}=\frac{8}{3}{e}^{i\cdot 0}=\frac{8}{3}.$$

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

Homework.

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

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

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

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

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

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

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

Represent the following complex numbers in exponential form:

4. $5-12i$.

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

5. $-3-4i$.

Answer: $5e^{i(\arctan \left(\frac{4}{3}\right)+\pi )}$.

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

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

7. $\sin \alpha +i(1-\cos \alpha )$.

Answer: $2\sin \frac{\pi }{2}{e}^{i\frac{\alpha }{2}}$.

8. Express the numbers $z_1$ and $z_2$ in exponential form and perform the specified operations:

$z_1^2\overline{z_2}$; $\frac{\overline{z_2}}{z_1}$, if $z_1=-\sqrt{3}+i\sqrt{2}$, $z_2=\sqrt{8}-\sqrt{8}$.

Answer: $16e^{i\frac{7\pi }{4}}$; $2e^{-i\frac{\pi }{2}}$.

Tags: complex numbers, exponential forms of a complex number, trigonometric forms of a complex number