Polar Form of Complex Numbers - Trigonometry

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The polar coordinates $\left ( r,\theta \right )$ of a point are $\left(2.1, $\frac{7\pi }{3}\right )$ . Convert these polar coordinates to rectangular coordinates.

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Given the polar coordinates $(r,\theta )$ , the -coordinate is $x= r \cos \theta$ . We can find this coordinate by substituting $r = 2.1, \theta = $\frac{7\pi }{3}$$ :

$x = r \cos \theta = 2.1 \cdot \cos $\frac{7\pi }{3}$ = 2.1 \cdot 0.5 = 1.05$

Likewise, given the polar coordinates $(r,\theta )$ , the -coordinate is $y= r \sin \theta$ . We can find this coordinate by substituting $r = 2.1, \theta = $\frac{7\pi }{3}$$ :

$y = r \sin \theta = 2.1 \cdot \sin $\frac{7\pi }{3}$ \approx 2.1 \cdot 0.8660 \approx 1.82$

Therefore the rectangular coordinates of the point $\left(2.1, $\frac{7\pi }{3}\right )$ are .

Express the complex number $z=4(\cos240^\circ+i\sin240^\circ)$ in rectangular form.

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To convert this number to rectangular form, first think about what $\cos240^\circ$ and $\sin240^\circ$ are equal to. Because $240^\circ=180^\circ+60^\circ$ , we can use a 30-60-90o reference triangle in the 3rd quadrant to determine these values.

$\cos240^\circ=-$\frac{1}{2}$$

$\sin240^\circ=-$\frac{\sqrt{3}$}{2}$

Now plug these in and continue solving:

$z=4(\cos240^\circ+i\sin240^\circ)$

$z=4(-$\frac{1}{2}$-i$\frac{\sqrt{3}$}{2})$

$z=-2-2$\sqrt{3}$i$

For the complex number , find the modulus and the angle $\theta=\tan^$-^1$\left ($\frac{y}{x}$ \right )$ . Then, express this number in polar form $z=r(\cos\theta+i\sin\theta)$ .

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This problem has given us formulas, so we just need to plug in and and solve.

$r=$\sqrt{218}$$

$\theta=\tan^$-^1$\left ($\frac{y}{x}$ \right )$

$\theta=\tan^$-^1$\left ($\frac{13}{7}$ \right )$

$\theta=61.7^\circ$

$z=r(\cos\theta+i\sin\theta)$

$z=$\sqrt{218}$(\cos61.7^\circ+i\sin61.7^\circ)$

Express the complex number $z=3(\cos30^\circ+i\sin30^\circ)$ in rectangular form .

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To convert this number to rectangular form, first think about what $\cos30^\circ$ and $\sin30^\circ$ are equal to. We can use a 30-60-90o reference triangle in the 1st quadrant to determine these values.

$\cos30^\circ=\frac{\sqrt{3}$}{2}$

$\sin30^\circ=\frac{1}{2}$$

Next, plug these values in and simplify:

$z=3(\cos30^\circ+i\sin30^\circ)$

$z=3($\frac{\sqrt{3}$}{2}+\frac{1}{2}$i)$

$z=\frac{3\sqrt{3}$}{2}+\frac{3}{2}$i$

For the complex number , find the modulus and the angle $\theta=\tan^$-^1$\left ($\frac{y}{x}$ \right )$ . Then, express this number in polar form $z=r(\cos\theta+i\sin\theta)$ .

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This problem has given us formulas, so we just need to plug in and and solve.

$r=$\sqrt{29}$$

$\theta=\tan^$-^1$\left ($\frac{y}{x}$ \right )$

$\theta=\tan^$-^1$\left ($\frac{-5}{2}$ \right )$

$\theta=-68.2^\circ$

$z=r(\cos\theta+i\sin\theta)$

$z=$\sqrt{29}$(\cos(-68.2^\circ)+i\sin(-68.2^\circ))$

Express the complex number in polar form.

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In order to complete this problem, you must understand three formulas that allow you to convert from the rectangular form of a complex number to the polar form of a complex number. These formulas are , $\theta=\tan^$-^1$\left ($\frac{y}{x}$ \right )$ , and the polar form $z=r(\cos\theta+i\sin\theta)$ . Additionally, understand that based on the given info, and . Begin by finding the modulus:

$r=$\sqrt{17}$$

Next, let's find the angle $\theta$ , also referred to as the amplitude of the complex number.

$\theta=\tan^$-^1$\left ($\frac{y}{x}$ \right )$

$\theta=\tan^$-^1$\left ($\frac{-1}{-4}$ \right )$

$\theta=\tan^$-^1$\left ($\frac{1}{4}$ \right )$

$\theta=14.04^\circ$

Finally, plug each of these into the polar form of a complex number:

$z=r(\cos\theta+i\sin\theta)$

$z=$\sqrt{17}$(\cos14.04^\circ+i\sin14.04^\circ)$

Multiply the following complex numbers (in polar form), giving the result in both polar and rectangular form.

$2(\cos70^\circ+i\sin70^\circ)\cdot 3(\cos20^\circ+i\sin20^\circ)$

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The modulus of the product of two complex numbers is the product of their moduli, and the amplitude of the product is the sum of their amplitudes.

Therefore, the new modulus will be $2\cdot 3=6$ and the new amplitude will be $70^\circ+20^\circ=90^\circ$ . Therefore

$2(\cos70^\circ+i\sin70^\circ)\cdot 3(\cos20^\circ+i\sin20^\circ)$ $=6(\cos90^\circ+i\sin90^\circ)$

We must also express this in rectangular form, which we can do by substituting $\cos90^\circ=0$ and $\sin90^\circ=1$ . We get:

$6(\cos90^\circ+i\sin90^\circ)$

Find the following quotients, given that $z_1=10(\cos40^\circ+i\sin40^\circ)$ and $z_2=2(\cos220^\circ-i\sin220^\circ)$ . Give results in both polar and rectangular forms.

(a) $\frac{z_1}{z_2}$

(b) $\frac{z_2}{z_1}$

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The modulus of the quotient of two complex numbers is the modulus of the dividend divided by the modulus of the divisor. The amplitude of the quotient is the amplitude of the dividend minus the amplitude of hte divisor.

(a) The modulus for $\frac{z_1}{z_2}$ is equal to $$\frac{10}{2}$=5$ . The amplitude for $\frac{z_1}{z_2}$ is equal to $40^\circ-220^\circ=-180^\circ=180^\circ$ . (We have chosen to represent this as the coterminal angle $180^\circ$ rather than $-180^\circ$ as it is more conventional to represent angle measures as a positive angle between $0^\circ$ and $360^\circ$ .) Putting this together, we get $$\frac{z_1}{z_2}$=5(\cos180^\circ+i\sin180^\circ)$ . To represent this in rectangular form, substitute $\cos180^\circ=-1$ and $\sin180^\circ=0$ to get $$\frac{z_1}{z_2}$=5(-1+0i)=-5$ .

(b) The modulus for $\frac{z_2}{z_1}$ is equal to $\frac{2}{10}$=\frac{1}{5}$ . The amplitude for $\frac{z_2}{z_1}$ is equal to $220^\circ-40^\circ=180^\circ$ . Putting this together, we get $$\frac{z_2}{z_1}$=\frac{1}{5}$(\cos180^\circ+i\sin180^\circ)$ . To represent this in rectangular form, substitute $\cos180^\circ=-1$ and $\sin180^\circ=0$ to get $\frac{z_2}{z_1}$=\frac{1}{5}$(-1+0i)=-$\frac{1}{5}$ .

The polar coordinates $\left ( r,\theta \right )$ of a point are $\left(2.1, $\frac{7\pi }{3}\right )$ . Convert these polar coordinates to rectangular coordinates.

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Given the polar coordinates $(r,\theta )$ , the -coordinate is $x= r \cos \theta$ . We can find this coordinate by substituting $r = 2.1, \theta = $\frac{7\pi }{3}$$ :

$x = r \cos \theta = 2.1 \cdot \cos $\frac{7\pi }{3}$ = 2.1 \cdot 0.5 = 1.05$

Likewise, given the polar coordinates $(r,\theta )$ , the -coordinate is $y= r \sin \theta$ . We can find this coordinate by substituting $r = 2.1, \theta = $\frac{7\pi }{3}$$ :

$y = r \sin \theta = 2.1 \cdot \sin $\frac{7\pi }{3}$ \approx 2.1 \cdot 0.8660 \approx 1.82$

Therefore the rectangular coordinates of the point $\left(2.1, $\frac{7\pi }{3}\right )$ are .

Express the complex number $z=4(\cos240^\circ+i\sin240^\circ)$ in rectangular form.

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To convert this number to rectangular form, first think about what $\cos240^\circ$ and $\sin240^\circ$ are equal to. Because $240^\circ=180^\circ+60^\circ$ , we can use a 30-60-90o reference triangle in the 3rd quadrant to determine these values.

$\cos240^\circ=-$\frac{1}{2}$$

$\sin240^\circ=-$\frac{\sqrt{3}$}{2}$

Now plug these in and continue solving:

$z=4(\cos240^\circ+i\sin240^\circ)$

$z=4(-$\frac{1}{2}$-i$\frac{\sqrt{3}$}{2})$

$z=-2-2$\sqrt{3}$i$

For the complex number , find the modulus and the angle $\theta=\tan^$-^1$\left ($\frac{y}{x}$ \right )$ . Then, express this number in polar form $z=r(\cos\theta+i\sin\theta)$ .

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This problem has given us formulas, so we just need to plug in and and solve.

$r=$\sqrt{218}$$

$\theta=\tan^$-^1$\left ($\frac{y}{x}$ \right )$

$\theta=\tan^$-^1$\left ($\frac{13}{7}$ \right )$

$\theta=61.7^\circ$

$z=r(\cos\theta+i\sin\theta)$

$z=$\sqrt{218}$(\cos61.7^\circ+i\sin61.7^\circ)$

Express the complex number $z=3(\cos30^\circ+i\sin30^\circ)$ in rectangular form .

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To convert this number to rectangular form, first think about what $\cos30^\circ$ and $\sin30^\circ$ are equal to. We can use a 30-60-90o reference triangle in the 1st quadrant to determine these values.

$\cos30^\circ=\frac{\sqrt{3}$}{2}$

$\sin30^\circ=\frac{1}{2}$$

Next, plug these values in and simplify:

$z=3(\cos30^\circ+i\sin30^\circ)$

$z=3($\frac{\sqrt{3}$}{2}+\frac{1}{2}$i)$

$z=\frac{3\sqrt{3}$}{2}+\frac{3}{2}$i$

For the complex number , find the modulus and the angle $\theta=\tan^$-^1$\left ($\frac{y}{x}$ \right )$ . Then, express this number in polar form $z=r(\cos\theta+i\sin\theta)$ .

Tap to see back →

This problem has given us formulas, so we just need to plug in and and solve.

$r=$\sqrt{29}$$

$\theta=\tan^$-^1$\left ($\frac{y}{x}$ \right )$

$\theta=\tan^$-^1$\left ($\frac{-5}{2}$ \right )$

$\theta=-68.2^\circ$

$z=r(\cos\theta+i\sin\theta)$

$z=$\sqrt{29}$(\cos(-68.2^\circ)+i\sin(-68.2^\circ))$

Express the complex number in polar form.

Tap to see back →

In order to complete this problem, you must understand three formulas that allow you to convert from the rectangular form of a complex number to the polar form of a complex number. These formulas are , $\theta=\tan^$-^1$\left ($\frac{y}{x}$ \right )$ , and the polar form $z=r(\cos\theta+i\sin\theta)$ . Additionally, understand that based on the given info, and . Begin by finding the modulus:

$r=$\sqrt{17}$$

Next, let's find the angle $\theta$ , also referred to as the amplitude of the complex number.

$\theta=\tan^$-^1$\left ($\frac{y}{x}$ \right )$

$\theta=\tan^$-^1$\left ($\frac{-1}{-4}$ \right )$

$\theta=\tan^$-^1$\left ($\frac{1}{4}$ \right )$

$\theta=14.04^\circ$

Finally, plug each of these into the polar form of a complex number:

$z=r(\cos\theta+i\sin\theta)$

$z=$\sqrt{17}$(\cos14.04^\circ+i\sin14.04^\circ)$

Multiply the following complex numbers (in polar form), giving the result in both polar and rectangular form.

$2(\cos70^\circ+i\sin70^\circ)\cdot 3(\cos20^\circ+i\sin20^\circ)$

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The modulus of the product of two complex numbers is the product of their moduli, and the amplitude of the product is the sum of their amplitudes.

Therefore, the new modulus will be $2\cdot 3=6$ and the new amplitude will be $70^\circ+20^\circ=90^\circ$ . Therefore

$2(\cos70^\circ+i\sin70^\circ)\cdot 3(\cos20^\circ+i\sin20^\circ)$ $=6(\cos90^\circ+i\sin90^\circ)$

We must also express this in rectangular form, which we can do by substituting $\cos90^\circ=0$ and $\sin90^\circ=1$ . We get:

$6(\cos90^\circ+i\sin90^\circ)$

Find the following quotients, given that $z_1=10(\cos40^\circ+i\sin40^\circ)$ and $z_2=2(\cos220^\circ-i\sin220^\circ)$ . Give results in both polar and rectangular forms.

(a) $\frac{z_1}{z_2}$

(b) $\frac{z_2}{z_1}$

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The modulus of the quotient of two complex numbers is the modulus of the dividend divided by the modulus of the divisor. The amplitude of the quotient is the amplitude of the dividend minus the amplitude of hte divisor.

(a) The modulus for $\frac{z_1}{z_2}$ is equal to $$\frac{10}{2}$=5$ . The amplitude for $\frac{z_1}{z_2}$ is equal to $40^\circ-220^\circ=-180^\circ=180^\circ$ . (We have chosen to represent this as the coterminal angle $180^\circ$ rather than $-180^\circ$ as it is more conventional to represent angle measures as a positive angle between $0^\circ$ and $360^\circ$ .) Putting this together, we get $$\frac{z_1}{z_2}$=5(\cos180^\circ+i\sin180^\circ)$ . To represent this in rectangular form, substitute $\cos180^\circ=-1$ and $\sin180^\circ=0$ to get $$\frac{z_1}{z_2}$=5(-1+0i)=-5$ .

(b) The modulus for $\frac{z_2}{z_1}$ is equal to $\frac{2}{10}$=\frac{1}{5}$ . The amplitude for $\frac{z_2}{z_1}$ is equal to $220^\circ-40^\circ=180^\circ$ . Putting this together, we get $$\frac{z_2}{z_1}$=\frac{1}{5}$(\cos180^\circ+i\sin180^\circ)$ . To represent this in rectangular form, substitute $\cos180^\circ=-1$ and $\sin180^\circ=0$ to get $\frac{z_2}{z_1}$=\frac{1}{5}$(-1+0i)=-$\frac{1}{5}$ .

The polar coordinates $\left ( r,\theta \right )$ of a point are $\left(2.1, $\frac{7\pi }{3}\right )$ . Convert these polar coordinates to rectangular coordinates.

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Given the polar coordinates $(r,\theta )$ , the -coordinate is $x= r \cos \theta$ . We can find this coordinate by substituting $r = 2.1, \theta = $\frac{7\pi }{3}$$ :

$x = r \cos \theta = 2.1 \cdot \cos $\frac{7\pi }{3}$ = 2.1 \cdot 0.5 = 1.05$

Likewise, given the polar coordinates $(r,\theta )$ , the -coordinate is $y= r \sin \theta$ . We can find this coordinate by substituting $r = 2.1, \theta = $\frac{7\pi }{3}$$ :

$y = r \sin \theta = 2.1 \cdot \sin $\frac{7\pi }{3}$ \approx 2.1 \cdot 0.8660 \approx 1.82$

Therefore the rectangular coordinates of the point $\left(2.1, $\frac{7\pi }{3}\right )$ are .

Express the complex number $z=4(\cos240^\circ+i\sin240^\circ)$ in rectangular form.

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To convert this number to rectangular form, first think about what $\cos240^\circ$ and $\sin240^\circ$ are equal to. Because $240^\circ=180^\circ+60^\circ$ , we can use a 30-60-90o reference triangle in the 3rd quadrant to determine these values.

$\cos240^\circ=-$\frac{1}{2}$$

$\sin240^\circ=-$\frac{\sqrt{3}$}{2}$

Now plug these in and continue solving:

$z=4(\cos240^\circ+i\sin240^\circ)$

$z=4(-$\frac{1}{2}$-i$\frac{\sqrt{3}$}{2})$

$z=-2-2$\sqrt{3}$i$

For the complex number , find the modulus and the angle $\theta=\tan^$-^1$\left ($\frac{y}{x}$ \right )$ . Then, express this number in polar form $z=r(\cos\theta+i\sin\theta)$ .

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This problem has given us formulas, so we just need to plug in and and solve.

$r=$\sqrt{218}$$

$\theta=\tan^$-^1$\left ($\frac{y}{x}$ \right )$

$\theta=\tan^$-^1$\left ($\frac{13}{7}$ \right )$

$\theta=61.7^\circ$

$z=r(\cos\theta+i\sin\theta)$

$z=$\sqrt{218}$(\cos61.7^\circ+i\sin61.7^\circ)$

Express the complex number $z=3(\cos30^\circ+i\sin30^\circ)$ in rectangular form .

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To convert this number to rectangular form, first think about what $\cos30^\circ$ and $\sin30^\circ$ are equal to. We can use a 30-60-90o reference triangle in the 1st quadrant to determine these values.

$\cos30^\circ=\frac{\sqrt{3}$}{2}$

$\sin30^\circ=\frac{1}{2}$$

Next, plug these values in and simplify:

$z=3(\cos30^\circ+i\sin30^\circ)$

$z=3($\frac{\sqrt{3}$}{2}+\frac{1}{2}$i)$

$z=\frac{3\sqrt{3}$}{2}+\frac{3}{2}$i$