Lesson Objectives
• Learn how to find products of complex numbers in polar form
• Learn how to find quotients of complex numbers in polar form

Products and Quotients of Complex Numbers in Polar Form

In the last lesson, we learned how to write a complex number in polar form (trigonometric form). Here, we will learn how to quickly multiply or divide two complex numbers in polar form using the product/quotient theorems. Let's begin by thinking about products of complex numbers in polar form.
Suppose we have the following two complex numbers written in polar form, where r is the absolute value of the complex number (magnitude) and θ is the argument (direction angle). $$r_{1}(\text{cos}\hspace{.1em}θ_{1}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1})$$ $$r_{2}(\text{cos}\hspace{.1em}θ_{2}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{2})$$ $$r_{1}(\text{cos}\hspace{.1em}θ_{1}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}) \cdot r_{2}(\text{cos}\hspace{.1em}θ_{2}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{2})$$ To begin, let's multiply the two absolute values: $$r_{1}\cdot r_{2}(\text{cos}\hspace{.1em}θ_{1}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}) (\text{cos}\hspace{.1em}θ_{2}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{2})$$ Now, let's use FOIL: $$r_{1}\cdot r_{2}[\text{cos}\hspace{.1em}θ_{1}\cdot \text{cos}\hspace{.1em}θ_{2}+ i \hspace{.1em}\text{cos}\hspace{.1em}θ_{1}\cdot \hspace{.1em}\text{sin}\hspace{.1em}θ_{2}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\cdot \text{cos}\hspace{.1em}θ_{2}+ i^{2}\hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\cdot \text{sin}\hspace{.1em}θ_{2}]$$ By definition $i^{2}=1$, let's replace this: $$r_{1}\cdot r_{2}[\text{cos}\hspace{.1em}θ_{1}\cdot \text{cos}\hspace{.1em}θ_{2}+ i \hspace{.1em}\text{cos}\hspace{.1em}θ_{1}\cdot \hspace{.1em}\text{sin}\hspace{.1em}θ_{2}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\cdot \text{cos}\hspace{.1em}θ_{2}- \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\cdot \text{sin}\hspace{.1em}θ_{2}]$$ Rearrange terms: $$r_{1}\cdot r_{2}[\text{cos}\hspace{.1em}θ_{1}\cdot \text{cos}\hspace{.1em}θ_{2}- \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\cdot \text{sin}\hspace{.1em}θ_{2}+ i \hspace{.1em}\text{cos}\hspace{.1em}θ_{1}\cdot \hspace{.1em}\text{sin}\hspace{.1em}θ_{2}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\cdot \text{cos}\hspace{.1em}θ_{2}]$$ Factor out the $i$: $$r_{1}\cdot r_{2}[(\text{cos}\hspace{.1em}θ_{1}\cdot \text{cos}\hspace{.1em}θ_{2}- \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\cdot \text{sin}\hspace{.1em}θ_{2}) + i (\hspace{.1em}\text{cos}\hspace{.1em}θ_{1}\cdot \hspace{.1em}\text{sin}\hspace{.1em}θ_{2}+ \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\cdot \text{cos}\hspace{.1em}θ_{2})]$$ Recall the sum identity for cosine: $$\text{cos}(A + B)=\text{cos}\hspace{.1em}A \hspace{.1em}\text{cos}B - \text{sin}\hspace{.1em}A \hspace{.1em}\text{sin}B$$ We can use this to rewrite: $$\text{cos}\hspace{.1em}θ_{1}\cdot \text{cos}\hspace{.1em}θ_{2}- \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\cdot \text{sin}\hspace{.1em}θ_{2}=\text{cos}(θ_{1}+ θ_{2})$$ Let's replace this: $$r_{1}\cdot r_{2}[\text{cos}(θ_{1}+ θ_{2}) + i (\hspace{.1em}\text{cos}\hspace{.1em}θ_{1}\cdot \hspace{.1em}\text{sin}\hspace{.1em}θ_{2}+ \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\cdot \text{cos}\hspace{.1em}θ_{2})]$$ Recall the sum identity for sine: $$\text{sin}(A + B)=\text{sin}\hspace{.1em}A \hspace{.1em}\text{cos}\hspace{.1em}B + \text{cos}\hspace{.1em}A \hspace{.1em}\text{sin}\hspace{.1em}B$$ We can use this to rewrite: $$\hspace{.1em}\text{cos}\hspace{.1em}θ_{1}\cdot \hspace{.1em}\text{sin}\hspace{.1em}θ_{2}+ \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\cdot \text{cos}\hspace{.1em}θ_{2}=\text{sin}(θ_{1}+ θ_{2})$$ Let's replace this: $$r_{1}\cdot r_{2}[\text{cos}(θ_{1}+ θ_{2}) + i \hspace{.1em}\text{sin}(θ_{1}+ θ_{2})]$$ The result gives us our product theorem for multiplying two complex numbers.

Product Theorem for Multiplying Two Complex Numbers

$$[r_{1}(\text{cos}\hspace{.1em}θ_{1}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1})] \cdot [r_{2}(\text{cos}\hspace{.1em}θ_{2}+ i \hspace{.1em} \text{sin}\hspace{.1em}θ_{2})]=r_{1}\cdot r_{2}[\text{cos}(θ_{1}+ θ_{2}) + i \hspace{.1em}\text{sin}(θ_{1}+ θ_{2})]$$ In compact form: $$(r_{1}\hspace{.1em}\text{cis}\hspace{.1em}θ_{1})(r_{2}\hspace{.1em}\text{cis}\hspace{.1em}θ_{2})=r_{1}\hspace{.1em}r_{2}\hspace{.1em}\text{cis}(θ_{1}+ θ_{2})$$ In other words, to multiply two complex numbers in polar form, multiply their absolute values and add their arguments. Let's look at a few examples.
Example #1: Simplify, write in rectangular form. $$3(\text{cos}\hspace{.1em}150° + i \hspace{.1em}\text{sin}\hspace{.1em}150°) \cdot 3 (\text{cos}\hspace{.1em}90° + i \hspace{.1em}\text{sin}\hspace{.1em}90°)$$ Let's reference our product theorem: $$[r_{1}(\text{cos}\hspace{.1em}θ_{1}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1})] \cdot [r_{2}(\text{cos}\hspace{.1em}θ_{2}+ i \hspace{.1em} \text{sin}\hspace{.1em}θ_{2})]=r_{1}\cdot r_{2}[\text{cos}(θ_{1}+ θ_{2}) + i \hspace{.1em}\text{sin}(θ_{1}+ θ_{2})]$$ $$3(\text{cos}\hspace{.1em}150° + i \hspace{.1em}\text{sin}\hspace{.1em}150°) \cdot 3 (\text{cos}\hspace{.1em}90° + i \hspace{.1em}\text{sin}\hspace{.1em}90°)$$ $$=3 \cdot 3[\text{cos}(150° + 90°) + i \hspace{.1em}\text{sin}(150° + 90°)]$$ $$=9(\text{cos}\hspace{.1em}240° + i \hspace{.1em}\text{sin}\hspace{.1em}240°)$$ $$=9\left(-\frac{1}{2}- \frac{i\sqrt{3}}{2}\right)$$ $$=-\frac{9}{2}- \frac{9\sqrt{3}}{2}i$$ Note: We can convert the original problem to rectangular form and show we get the same answer: $$3(\text{cos}\hspace{.1em}150° + i \hspace{.1em}\text{sin}\hspace{.1em}150°) \cdot 3 (\text{cos}\hspace{.1em}90° + i \hspace{.1em}\text{sin}\hspace{.1em}90°)$$ $$=\left(-\frac{3\sqrt{3}}{2}+ \frac{3}{2}i\right)(3i)$$ $$=\left(-\frac{9\sqrt{3}}{2}i + \frac{9}{2}i^{2}\right)$$ $$=\left(-\frac{9\sqrt{3}}{2}i - \frac{9}{2}\right)$$ Put in standard form a + bi: $$=- \frac{9}{2}-\frac{9\sqrt{3}}{2}i$$ Example #2: Simplify, write in polar form. $$(-2\sqrt{2}- 2i\sqrt{2})(-\sqrt{3}- i)$$ Let's write each complex number in polar form, then use our product theorem. Note, this is just for the purposes of our tutorial, it might be faster here to use FOIL and then convert the answer to polar form. $$-2\sqrt{2}- 2i\sqrt{2}$$ Since the real part is negative and the imaginary part is negative, we know this complex number lies in quadrant III. $$r=\sqrt{(-2\sqrt{2})^{2}+ (-2\sqrt{2})^{2}}$$ $$r=\sqrt{16}$$ $$r=4$$ $$\text{tan}\hspace{.1em}θ=\frac{-2\sqrt{2}}{-2\sqrt{2}}=1$$ Our reference angle: $$\text{tan}^{-1}(1)=45°$$ In quadrant III our angle with a reference angle of 45°: $$180° + 45°=225°$$ $$-2\sqrt{2}- 2i\sqrt{2}=4(\text{cos}\hspace{.1em}225° + i \hspace{.1em}\text{sin}\hspace{.1em}225°)$$ Now, let's find the polar form of our other complex number: $$-\sqrt{3}- i$$ Again, since the real part is negative and the imaginary part is negative, we know this complex number lies in quadrant III. $$r=\sqrt{(-\sqrt{3})^2 + (-1)^2}$$ $$r=\sqrt{3 + 1}$$ $$r=\sqrt{4}$$ $$r=2$$ $$\text{tan}\hspace{.1em}θ=\frac{-1}{-\sqrt{3}}=\frac{\sqrt{3}}{3}$$ Our reference angle: $$\text{tan}^{-1}\left(\frac{\sqrt{3}}{3}\right)=30°$$ In quadrant III our angle with a reference angle of 30°: $$180° + 30°=210°$$ $$-\sqrt{3}- i=2(\text{cos}\hspace{.1em}210° + i \hspace{.1em}\text{sin}210°)$$ Now, let's return to the original problem: $$(-2\sqrt{2}- 2i\sqrt{2})(-\sqrt{3}- i)$$ $$=4(\text{cos}\hspace{.1em}225° + i \hspace{.1em}\text{sin}\hspace{.1em}225°) \cdot 2(\text{cos}\hspace{.1em}210° + i \hspace{.1em}\text{sin}210°)$$ Let's reference our product theorem: $$[r_{1}(\text{cos}\hspace{.1em}θ_{1}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1})] \cdot [r_{2}(\text{cos}\hspace{.1em}θ_{2}+ i \hspace{.1em} \text{sin}\hspace{.1em}θ_{2})]=r_{1}\cdot r_{2}[\text{cos}(θ_{1}+ θ_{2}) + i \hspace{.1em}\text{sin}(θ_{1}+ θ_{2})]$$ $$4(\text{cos}\hspace{.1em}225° + i \hspace{.1em}\text{sin}\hspace{.1em}225°) \cdot 2(\text{cos}\hspace{.1em}210° + i \hspace{.1em}\text{sin}210°)$$ $$=4 \cdot 2[\text{cos}(225° + 210°) + i \hspace{.1em}\text{sin}(225° + 210°)]$$ $$=8 (\text{cos}\hspace{.1em}435° + i \hspace{.1em}\text{sin}\hspace{.1em}435°)$$ Although it isn't necessary, some teachers want an angle between 0 (inclusive) and 360° (exclusive). Here, we can just find the coterminal angle: $$435° - 360°=75°$$ $$8 (\text{cos}\hspace{.1em}435° + i \hspace{.1em}\text{sin}\hspace{.1em}435°)$$ $$=8 (\text{cos}\hspace{.1em}75° + i \hspace{.1em}\text{sin}\hspace{.1em}75°)$$

Dividing Complex Numbers in Polar Form

Let's again return to our two complex numbers written in polar form: $$r_{1}(\text{cos}\hspace{.1em}θ_{1}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1})$$ $$r_{2}(\text{cos}\hspace{.1em}θ_{2}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{2})$$ Suppose we want to find the quotient: $$\frac{r_{1}(\text{cos}\hspace{.1em}θ_{1}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1})}{r_{2}(\text{cos}\hspace{.1em}θ_{2}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{2})}$$ Let's begin by multiplying the numerator and denominator by the complex conjugate of the denominator: $$\frac{r_{1}(\text{cos}\hspace{.1em}θ_{1}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1})}{r_{2}(\text{cos}\hspace{.1em}θ_{2}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{2})}\cdot \frac{r_{2}(\text{cos}\hspace{.1em}θ_{2}- i \hspace{.1em}\text{sin}\hspace{.1em}θ_{2})}{r_{2}(\text{cos}\hspace{.1em}θ_{2}- i \hspace{.1em}\text{sin}\hspace{.1em}θ_{2})}$$ Let's work on the numerator first: $$r_{1}(\text{cos}\hspace{.1em}θ_{1}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}) \cdot r_{2}(\text{cos}\hspace{.1em}θ_{2}- i \hspace{.1em}\text{sin}\hspace{.1em}θ_{2})$$ To begin, let's multiply the two absolute values: $$r_{1}(\text{cos}\hspace{.1em}θ_{1}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}) \cdot r_{2}(\text{cos}\hspace{.1em}θ_{2}- i \hspace{.1em}\text{sin}\hspace{.1em}θ_{2})$$ $$=r_{1}\cdot r_{2}(\text{cos}\hspace{.1em}θ_{1}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1})(\text{cos}\hspace{.1em}θ_{2}- i \hspace{.1em}\text{sin}\hspace{.1em}θ_{2})$$ Now, let's use FOIL: $$=r_{1}\cdot r_{2}(\text{cos}\hspace{.1em}θ_{1}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1})(\text{cos}\hspace{.1em}θ_{2}- i \hspace{.1em}\text{sin}\hspace{.1em}θ_{2})$$ $$=r_{1}\cdot r_{2}[(\text{cos}\hspace{.1em}θ_{1}\cdot \text{cos}\hspace{.1em}θ_{2}- i \hspace{.1em}\text{cos}\hspace{.1em}θ_{1}\cdot \hspace{.1em}\text{sin}\hspace{.1em}θ_{2}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\hspace{.1em}\text{cos}\hspace{.1em}θ_{2}- i^{2}\hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\cdot \text{sin}\hspace{.1em}θ_{2})]$$ By definition $i^{2}=1$, let's replace this: $$=r_{1}\cdot r_{2}[(\text{cos}\hspace{.1em}θ_{1}\cdot \text{cos}\hspace{.1em}θ_{2}- i \hspace{.1em}\text{cos}\hspace{.1em}θ_{1}\cdot \hspace{.1em}\text{sin}\hspace{.1em}θ_{2}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\hspace{.1em}\text{cos}\hspace{.1em}θ_{2}+ \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\cdot \text{sin}\hspace{.1em}θ_{2})]$$ Rearrange terms: $$r_{1}\cdot r_{2}[(\text{cos}\hspace{.1em}θ_{1}\cdot \text{cos}\hspace{.1em}θ_{2}+ \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\cdot \text{sin}\hspace{.1em}θ_{2}- i \hspace{.1em}\text{cos}\hspace{.1em}θ_{1}\cdot \hspace{.1em}\text{sin}\hspace{.1em}θ_{2}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\cdot \text{cos}\hspace{.1em}θ_{2})]$$ Factor out the $i$: $$r_{1}\cdot r_{2}[(\text{cos}\hspace{.1em}θ_{1}\cdot \text{cos}\hspace{.1em}θ_{2}+ \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\cdot \text{sin}\hspace{.1em}θ_{2}) + i (-\hspace{.1em}\text{cos}\hspace{.1em}θ_{1}\cdot \hspace{.1em}\text{sin}\hspace{.1em}θ_{2}+ \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\cdot \text{cos}\hspace{.1em}θ_{2})]$$ Consider the following difference identities for sine and cosine: $$\text{cos}(A - B)=\text{cos}\hspace{.1em}A \hspace{.1em}\text{cos}\hspace{.1em}B + \text{sin}\hspace{.1em}A \hspace{.1em}\text{sin}\hspace{.1em}B$$ $$\text{cos}\hspace{.1em}θ_{1}\cdot \text{cos}\hspace{.1em}θ_{2}+ \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\cdot \text{sin}\hspace{.1em}θ_{2}=\text{cos}(θ_{1}- θ_{2})$$ $$\text{sin}(A - B)=\text{sin}\hspace{.1em}A \hspace{.1em}\text{cos}\hspace{.1em}B - \text{cos}\hspace{.1em}A \hspace{.1em}\text{sin}\hspace{.1em}B$$ $$-\hspace{.1em}\text{cos}\hspace{.1em}θ_{1}\cdot \hspace{.1em}\text{sin}\hspace{.1em}θ_{2}+ \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\cdot \text{cos}\hspace{.1em}θ_{2}$$ $$=\hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\cdot \text{cos}\hspace{.1em}θ_{2}-\hspace{.1em}\text{cos}\hspace{.1em}θ_{1}\cdot \hspace{.1em}\text{sin}\hspace{.1em}θ_{2}=\text{sin}(θ_{1}- θ_{2})$$ Let's use these identities to rewrite our numerator: $$r_{1}\cdot r_{2}[(\text{cos}\hspace{.1em}θ_{1}\cdot \text{cos}\hspace{.1em}θ_{2}+ \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\cdot \text{sin}\hspace{.1em}θ_{2}) + i (-\hspace{.1em}\text{cos}\hspace{.1em}θ_{1}\cdot \hspace{.1em}\text{sin}\hspace{.1em}θ_{2}+ \hspace{.1em}\text{sin}\hspace{.1em}θ_{1}\cdot \text{cos}\hspace{.1em}θ_{2})]$$ $$=r_{1}\cdot r_{2}[\text{cos}(θ_{1}- θ_{2}) + i \hspace{.1em}\text{sin}(θ_{1}- θ_{2})]$$ Let's now work on the denominator: $$r_{2}(\text{cos}\hspace{.1em}θ_{2}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{2}) \cdot r_{2}(\text{cos}\hspace{.1em}θ_{2}- i \hspace{.1em}\text{sin}\hspace{.1em}θ_{2})$$ Multiply the two absolute values together: $$(r_{2})^{2}(\text{cos}\hspace{.1em}θ_{2}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{2})(\text{cos}\hspace{.1em}θ_{2}- i \hspace{.1em}\text{sin}\hspace{.1em}θ_{2})$$ Use the formula for conjugates: $$(r_{2})^{2}(\text{cos}^{2}\hspace{.1em}θ_{2}- i^2 \hspace{.1em}\text{sin}^{2}\hspace{.1em}θ_{2})$$ By definition $i^{2}$=1, let's replace this: $$(r_{2})^{2}(\text{cos}^{2}\hspace{.1em}θ_{2}+ \hspace{.1em}\text{sin}^{2}\hspace{.1em}θ_{2})$$ Use the Pythagorean Identity: $$\text{sin}^{2}\hspace{.1em}θ_{2}+ \text{cos}^2 \hspace{.1em}θ_{2}=1$$ $$(r_{2})^{2}(1)=(r_{2})^{2}$$ Let's now set up our numerator and denominator: $$\frac{r_{1}\cdot r_{2}[\text{cos}(θ_{1}- θ_{2}) + i \hspace{.1em}\text{sin}(θ_{1}- θ_{2})]}{(r_{2})^{2}}$$ We can remove a factor of r2 from the numerator and the denominator: $$\frac{r_{1}[\text{cos}(θ_{1}- θ_{2}) + i \hspace{.1em}\text{sin}(θ_{1}- θ_{2})]}{r_{2}}$$ Let's rewrite this as: $$\frac{r_{1}}{r_{2}}[\text{cos}(θ_{1}- θ_{2}) + i \hspace{.1em}\text{sin}(θ_{1}- θ_{2})]$$ The result gives us our quotient theorem for dividing two complex numbers.

Quotient Theorem

$$\frac{r_{1}(\text{cos}\hspace{.1em}θ_{1}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1})}{r_{2}(\text{cos}\hspace{.1em}θ_{2}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{2})}=\frac{r_{1}}{r_{2}}[\text{cos}(θ_{1}- θ_{2}) + i \hspace{.1em}\text{sin}(θ_{1}- θ_{2})]$$ In compact form: $$\frac{r_{1}\hspace{.1em}\text{cis}θ_{1}}{r_{2}\hspace{.1em}\text{cis}θ_{2}}=\frac{r_{1}}{r_{2}}\hspace{.1em}\text{cis}(θ_{1}- θ_{2})$$ In other words, to divide two complex numbers in polar form, divide their absolute values and subtract their arguments. Let's look at a few examples.
Example #3: Find the quotient, write the answer in rectangular form. $$\frac{\sqrt{30}\left(\text{cos}\hspace{.1em}\large{\frac{5π}{3}}+ i \hspace{.1em}\text{sin}\hspace{.1em}\large{\frac{5π}{3}}\right)}{3\left(\text{cos}\hspace{.1em}\large{\frac{7π}{4}}+ i \hspace{.1em}\text{sin}\hspace{.1em}\large{\frac{7π}{4}}\right)}$$ Let's reference our quotient theorem: $$\frac{r_{1}(\text{cos}\hspace{.1em}θ_{1}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1})}{r_{2}(\text{cos}\hspace{.1em}θ_{2}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{2})}=\frac{r_{1}}{r_{2}}[\text{cos}(θ_{1}- θ_{2}) + i \hspace{.1em}\text{sin}(θ_{1}- θ_{2})]$$ $$\frac{\sqrt{30}\left(\text{cos}\hspace{.1em}\large{\frac{5π}{3}}+ i \hspace{.1em}\text{sin}\hspace{.1em}\large{\frac{5π}{3}}\right)}{3\left(\text{cos}\hspace{.1em}\large{\frac{7π}{4}}+ i \hspace{.1em}\text{sin}\hspace{.1em}\large{\frac{7π}{4}}\right)}$$ $$=\frac{\sqrt{30}}{3}\left(\text{cos}\left(\frac{5π}{3}- \frac{7π}{4}\right) + i \hspace{.1em}\text{sin}\left(\frac{5π}{3}- \frac{7π}{4}\right)\right)$$ $$=\frac{\sqrt{30}}{3}\left(\text{cos}\left(\frac{20π}{12}- \frac{21π}{12}\right) + i \hspace{.1em}\text{sin}\left(\frac{20π}{12}- \frac{21π}{12}\right)\right)$$ $$=\frac{\sqrt{30}}{3}\left(\text{cos}\left(- \frac{π}{12}\right) + i \hspace{.1em}\text{sin}\left(- \frac{π}{12}\right)\right)$$ Let's add 2$π$ to find the coterminal angle: $$-\frac{π}{12}+ \frac{24π}{12}=\frac{23π}{12}$$ $$\frac{\sqrt{30}}{3}\left(\text{cos}\left(- \frac{π}{12}\right) + i \hspace{.1em}\text{sin}\left(- \frac{π}{12}\right)\right)$$ $$=\frac{\sqrt{30}}{3}\left(\text{cos}\left(\frac{23π}{12}\right) + i \hspace{.1em}\text{sin}\left(\frac{23π}{12}\right)\right)$$ Let's now write this in rectangular form: Recall, we can use our sum and difference identities for sine and cosine to find exact values: $$\text{cos}\left(\frac{23π}{12}\right)=\frac{\sqrt{6}+ \sqrt{2}}{4}$$ $$\text{sin}\left(\frac{23π}{12}\right)=\frac{\sqrt{2}- \sqrt{6}}{4}$$ Let's now write our rectangular form: $$\frac{\sqrt{30}}{3}\left(\text{cos}\left(\frac{23π}{12}\right) + i \hspace{.1em}\text{sin}\left(\frac{23π}{12}\right)\right)$$ $$=\frac{\sqrt{30}}{3}\cdot \frac{\sqrt{6}+ \sqrt{2}}{4}+ \frac{\sqrt{30}}{3}\cdot \frac{\sqrt{2}- \sqrt{6}}{4}i$$ $$=\frac{\sqrt{180}+ \sqrt{60}}{12}+ \frac{\sqrt{60}- \sqrt{180}}{12}i$$ $$=\frac{6\sqrt{5}+ 2\sqrt{15}}{12}+ \frac{2\sqrt{15}- 6\sqrt{5}}{12}i$$ $$=\frac{3\sqrt{5}+ \sqrt{15}}{6}+ \frac{\sqrt{15}- 3\sqrt{5}}{6}i$$ Example #4: Find the quotient, write the answer in polar form. $$\frac{1 - i\sqrt{3}}{\large{-\frac{\sqrt{14}}{2}- \frac{\sqrt{14}}{2}i}}$$ Let's first write each complex number in polar form, then use our quotient theorem. $$1 - i\sqrt{3}$$ Since the real part is positive and the imaginary part is negative, we know this complex number lies in quadrant IV. $$r=\sqrt{1^{2}+ (-\sqrt{3})^2}$$ $$r=\sqrt{1 + 3}$$ $$r=\sqrt{4}$$ $$r=2$$ $$\text{tan}\hspace{.1em}θ=\frac{-\sqrt{3}}{1}=-\sqrt{3}$$ Our reference angle: $$\text{tan}^{-1}(\sqrt{3})=60°$$ In quadrant IV our angle with a reference angle of 60°: $$360° - 60°=300°$$ $$1 - i \sqrt{3}=2(\text{cos}\hspace{.1em}300° + i \hspace{.1em}\text{sin}\hspace{.1em}300°)$$ Now, let's find our other complex number: $$-\frac{\sqrt{14}}{2}- \frac{\sqrt{14}}{2}i$$ Since the real part and the imaginary part are negative, we know this complex number lies in quadrant III. $$r=\sqrt{\left(-\frac{\sqrt{14}}{2}\right)^2 + \left(-\frac{\sqrt{14}}{2}\right)^2}$$ $$r=\sqrt{\frac{14}{4}+ \frac{14}{4}}$$ $$r=\sqrt{\frac{28}{4}}$$ $$r=\sqrt{7}$$ $$\text{tan}\hspace{.1em}θ=-\frac{\sqrt{14}}{2}\cdot -\frac{2}{\sqrt{14}}=1$$ Our reference angle: $$\text{tan}^{-1}(1)=45°$$ In quadrant III our angle with a reference angle of 45°: $$180° + 45°=225°$$ $$-\frac{\sqrt{14}}{2}- \frac{\sqrt{14}}{2}i=\sqrt{7}(\text{cos}\hspace{.1em}225° + i \hspace{.1em}\text{sin}\hspace{.1em}225°)$$ Now, let's return to the original problem: $$\frac{1 - i\sqrt{3}}{\large{-\frac{\sqrt{14}}{2}- \frac{\sqrt{14}}{2}i}}$$ $$=\frac{2(\text{cos}\hspace{.1em}300° + i \hspace{.1em}\text{sin}\hspace{.1em}300°)}{\sqrt{7}(\text{cos}\hspace{.1em}225° + i \hspace{.1em}\text{sin}\hspace{.1em}225°)}$$ Let's reference our quotient theorem: $$\frac{r_{1}(\text{cos}\hspace{.1em}θ_{1}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1})}{r_{2}(\text{cos}\hspace{.1em}θ_{2}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{2})}=\frac{r_{1}}{r_{2}}[\text{cos}(θ_{1}- θ_{2}) + i \hspace{.1em}\text{sin}(θ_{1}- θ_{2})]$$ $$\frac{2(\text{cos}\hspace{.1em}300° + i \hspace{.1em}\text{sin}\hspace{.1em}300°)}{\sqrt{7}(\text{cos}\hspace{.1em}225° + i \hspace{.1em}\text{sin}\hspace{.1em}225°)}$$ $$=\frac{2}{\sqrt{7}}(\text{cos}(300° - 225°) + i \hspace{.1em}\text{sin}(300° - 225°)$$ $$=\frac{2\sqrt{7}}{7}(\text{cos}75° + i \hspace{.1em}\text{sin}75°)$$

Skills Check:

Example #1

Simplify, write in rectangular form. $$4\hspace{.1em}\text{cis}\hspace{.1em}300° \cdot 2\hspace{.1em}\text{cis}\hspace{.1em}30°$$

A
$$18 - 18i\sqrt{3}$$
B
$$-15\sqrt{3}- 15i$$
C
$$15 - 15i\sqrt{3}$$
D
$$4\sqrt{3}- 4i$$
E
$$-3 - i\sqrt{5}$$

Example #2

Simplify, write in rectangular form. $$6\hspace{.1em}\text{cis}\hspace{.1em}120° \cdot 6\hspace{.1em}\text{cis}\hspace{.1em}120°$$

A
$$-9 + 9i\sqrt{3}$$
B
$$-18 - 18i\sqrt{3}$$
C
$$12\sqrt{6}- 12i\sqrt{2}$$
D
$$-6\sqrt{2}- 6i\sqrt{2}$$
E
$$-4 - 7i\sqrt{3}$$

Example #3

Simplify, write in rectangular form. $$\frac{\sqrt{10}\hspace{.1em}\text{cis}\hspace{.1em}300°}{\sqrt{17}\hspace{.1em}\text{cis}\hspace{.1em}270°}$$

A
$$\frac{6}{5}$$
B
$$\frac{\sqrt{15}}{8}+ \frac{3\sqrt{5}}{8}i$$
C
$$\frac{\sqrt{510}}{34}+ \frac{\sqrt{170}}{34}i$$
D
$$4 + i\sqrt{3}$$
E
$$\frac{\sqrt{51}}{3}+ \frac{\sqrt{19}}{3}i$$

Congrats, Your Score is 100%

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