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Trigonometry : Polar Form of Complex Numbers

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Example Questions

Example Question #1 : Polar Form Of Complex Numbers

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.

Possible Answers:

(2.1, 7.33)

(1.82, 1.05)

(1.05, 1.82)

(7.33, 2.1)

Correct answer:

(1.05, 1.82)

Explanation:

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 (1.05, 1.82) .

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Example Question #2 : Polar Form Of Complex Numbers

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

Possible Answers:

$z=-4-2\sqrt{3}i$

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

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

$z=4+i\sqrt{3}$

Correct answer:

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

Explanation:

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 30-60-90^o reference triangle in the 3rd quadrant to determine these values.

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

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

Now plug these in and continue solving:

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

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

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

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Example Question #3 : Polar Form Of Complex Numbers

For the complex number z=7+13i , find the modulus $r=\sqrt{x^2+y^2}$ and the angle $\theta=\tan^-^1\left (\frac{y}{x} \right )$ . Then, express this number in polar form $z=r(\cos\theta+i\sin\theta)$ .

Possible Answers:

$r=2\sqrt{30}$

$\theta=61.7^\circ$

$z=2\sqrt{30}(\cos61.7^\circ+i\sin61.7^\circ)$

$r=2\sqrt{30}$

$\theta=28.3^\circ$

$z=2\sqrt{30}(\cos28.3^\circ+i\sin28.3^\circ)$

$r=\sqrt{218}$

$\theta=28.3^\circ$

$z=\sqrt{218}(\cos28.3^\circ+i\sin28.3^\circ)$

$r=\sqrt{218}$

$\theta=61.7^\circ$

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

Correct answer:

$r=\sqrt{218}$

$\theta=61.7^\circ$

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

Explanation:

This problem has given us formulas, so we just need to plug in x=7 and y=13 and solve.

$r=\sqrt{x^2+y^2}$

$r=\sqrt{7^2+13^2}$

$r=\sqrt{218}$

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

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

$\theta=61.7^\circ$

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

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

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Example Question #4 : Polar Form Of Complex Numbers

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

Next, plug these values in and simplify:

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

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

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

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Example Question #5 : Polar Form Of Complex Numbers

For the complex number z=2-5i , find the modulus $r=\sqrt{x^2+y^2}$ and the angle $\theta=\tan^-^1\left (\frac{y}{x} \right )$ . Then, express this number in polar form $z=r(\cos\theta+i\sin\theta)$ .

Possible Answers:

r=5

$\theta=21.8^\circ$

$z=5(\cos21.8^\circ+i\sin21.8^\circ)$

$r=\sqrt{29}$

$\theta=-68.2^\circ$

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

$r=\sqrt{29}$

$\theta=21.8^\circ$

$z=\sqrt{29}(\cos21.8^\circ+i\sin21.8^\circ)$

r=5

$\theta=-68.2^\circ$

$z=5(\cos(-68.2^\circ)+i\sin(-68.2^\circ))$

Correct answer:

$r=\sqrt{29}$

$\theta=-68.2^\circ$

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

Explanation:

This problem has given us formulas, so we just need to plug in x=2 and y=-5 and solve.

$r=\sqrt{x^2+y^2}$

$r=\sqrt{2^2+(-5)^2}$

$r=\sqrt{29}$

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

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

$\theta=-68.2^\circ$

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

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

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Example Question #6 : Polar Form Of Complex Numbers

Express the complex number z=-1-4i in polar form.

Possible Answers:

$r=\sqrt{17}$

$\theta=75.96^\circ$

$z=\sqrt{17}(\cos75.96^\circ+i\sin75.96^\circ)$

$r=\sqrt{17}$

$\theta=14.04^\circ$

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

$r=\sqrt{15}$

$\theta=75.96^\circ$

$z=\sqrt{15}(\cos75.96^\circ+i\sin75.96^\circ)$

$r=\sqrt{15}$

$\theta=14.04^\circ$

$z=\sqrt{15}(\cos14.04^\circ+i\sin14.04^\circ)$

Correct answer:

$r=\sqrt{17}$

$\theta=14.04^\circ$

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

Explanation:

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 $r=\sqrt{x^2+y^2}$ , $\theta=\tan^-^1\left (\frac{y}{x} \right )$ , and the polar form $z=r(\cos\theta+i\sin\theta)$ . Additionally, understand that based on the given info, x=-1 and y=-4 . Begin by finding the modulus:

$r=\sqrt{x^2+y^2}$

$r=\sqrt{(-1)^2+(-4)^2}$

$r=\sqrt{17}$

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:

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

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

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Example Question #7 : Polar Form Of Complex Numbers

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

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

Possible Answers:

$5(\cos90^\circ+i\sin90^\circ)$ or

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

$5(\cos90^\circ+i\sin90^\circ)$ or

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

Correct answer:

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

Explanation:

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:

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

=6(0+1i)

=6i

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Example Question #8 : Polar Form Of Complex Numbers

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) $\frac{z_2}{z_1}$

Possible Answers:

(a) $\frac{z_1}{z_2}=5(\cos180^\circ+i\sin180^\circ)$ or $\frac{z_1}{z_2}=-5$

(b) $\frac{z_2}{z_1}=\frac{1}{5}(\cos180^\circ+i\sin180^\circ)$ or $\frac{z_2}{z_1}=-\frac{1}{5}$

(a) $\frac{z_1}{z_2}=\frac{1}{5}(\cos180^\circ+i\sin180^\circ)$ or $\frac{z_1}{z_2}=-5$

(b) $\frac{z_2}{z_1}=5(\cos180^\circ+i\sin180^\circ)$ or $\frac{z_2}{z_1}=-\frac{1}{5}$

(a) $\frac{z_1}{z_2}=5(\cos180^\circ+i\sin180^\circ)$ or $\frac{z_1}{z_2}=-5i$

(b) $\frac{z_2}{z_1}=\frac{1}{5}(\cos180^\circ+i\sin180^\circ)$ or $\frac{z_2}{z_1}=-\frac{1}{5}i$

(a) $\frac{z_1}{z_2}=\frac{1}{5}(\cos180^\circ+i\sin180^\circ)$ or $\frac{z_1}{z_2}=-5$

(b) $\frac{z_2}{z_1}=5(\cos180^\circ+i\sin180^\circ)$ or $\frac{z_2}{z_1}=-\frac{1}{5}$

Correct answer:

(a) $\frac{z_1}{z_2}=5(\cos180^\circ+i\sin180^\circ)$ or $\frac{z_1}{z_2}=-5$

(b) $\frac{z_2}{z_1}=\frac{1}{5}(\cos180^\circ+i\sin180^\circ)$ or $\frac{z_2}{z_1}=-\frac{1}{5}$

Explanation:

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 $\frac{z_2}{z_1}=\frac{1}{5}(-1+0i)=-\frac{1}{5}$ .

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Trigonometry : Polar Form of Complex Numbers

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