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Trigonometry : Trigonometry

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6 Diagnostic Tests 155 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Complex Numbers/Polar Form

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

Possible Answers:

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

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

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

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

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 : Complex Numbers/Polar Form

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:

$r=\sqrt{218}$

$\theta=61.7^\circ$

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

$r=2\sqrt{30}$

$\theta=61.7^\circ$

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

$r=\sqrt{218}$

$\theta=28.3^\circ$

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

$r=2\sqrt{30}$

$\theta=28.3^\circ$

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

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 #11 : Complex Numbers/Polar Form

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

Possible Answers:

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

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

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

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

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 #12 : Complex Numbers/Polar Form

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

$\theta=21.8^\circ$

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

$r=\sqrt{29}$

$\theta=21.8^\circ$

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

r=5

$\theta=-68.2^\circ$

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

$r=\sqrt{29}$

$\theta=-68.2^\circ$

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

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 #13 : Complex Numbers/Polar Form

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

Possible Answers:

$r=\sqrt{15}$

$\theta=14.04^\circ$

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

$r=\sqrt{15}$

$\theta=75.96^\circ$

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

$r=\sqrt{17}$

$\theta=14.04^\circ$

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

$r=\sqrt{17}$

$\theta=75.96^\circ$

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

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

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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Example Question #1 : Complex Numbers

Name the real part of this expression and the imaginary part of this expression: 3-7i .

Possible Answers:

Real:

Imaginary:

All parts are imaginary; there are no real parts

Real:

Imaginary: -7i

Real:

Imaginary:

All parts are real; there are no imaginary parts

Correct answer:

Real:

Imaginary: -7i

Explanation:

The real part of this expression includes any terms that do not have attached to them. Therefore the real part of this expression is 3. The imaginary part of this expression includes any terms with that cannot be further reduced; the imaginary part of this expression is -7i .

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Example Question #1 : Complex Numbers

Find the product of the complex number and its conjugate:

4+2i

Possible Answers:

12+16i

4-2i

Correct answer:

Explanation:

To solve this problem, we must first identify the conjugate of this complex number. The conjugate keeps the real portion of the number the same, but changes the sign of the imaginary part of the number. Therefore the conjugate of 4+2i is 4-2i . Now, we need to multiply these together using distribution, combining like terms, and substituting i^2=-1 .

$\left ( 4+2i \right )\left ( 4-2i \right )$

$16-8i+8i-4i^{2}$

16+4

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Example Question #3 : Complex Numbers

What is the complex conjugate of $\sqrt{3}-22i$ ?

Possible Answers:

-22i

$\sqrt{3}+22i$

$\sqrt{3-22i}$

$\sqrt{3}-22i$

$\sqrt{3+22i}$

Correct answer:

$\sqrt{3}+22i$

Explanation:

To solve this problem, we must understand what a complex conjugate is and how it relates to a complex number. The conjugate of a number a+bi is a-bi . Therefore the conjugate of $\sqrt{3}-22i$ is $\sqrt{3}+22i$ .

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Trigonometry : Trigonometry

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #1 : Complex Numbers/Polar Form

Example Question #3 : Complex Numbers/Polar Form

Example Question #11 : Complex Numbers/Polar Form

Example Question #12 : Complex Numbers/Polar Form

Example Question #13 : Complex Numbers/Polar Form

Example Question #1 : Polar Form Of Complex Numbers

Example Question #1 : Polar Form Of Complex Numbers

Example Question #1 : Complex Numbers

Example Question #1 : Complex Numbers

Example Question #3 : Complex Numbers

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