Product and Quotient Theorems Lesson

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: $$3(\text{cos}\hspace{.1em}150° + i \hspace{.1em}\text{sin}\hspace{.1em}150°) \cdot 3 (\text{cos}\hspace{.1em}90° + i \hspace{.1em}\text{sin}\hspace{.1em}90°)$$ $$=\left(-\frac{3\sqrt{3}}{2}+ \frac{3}{2}i\right)(3i)$$ $$=-\frac{9\sqrt{3}}{2}i + \frac{9}{2}i^{2}$$ $$=-\frac{9\sqrt{3}}{2}i - \frac{9}{2}$$ 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. $$-2\sqrt{2}- 2i\sqrt{2}$$ 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°: $$180° + 30°=210°$$ $$-\sqrt{3}- i=2(\text{cos}\hspace{.1em}210° + i \hspace{.1em}\text{sin}\hspace{.1em}210°)$$ Now, let's return to the original problem: $$(-2\sqrt{2}- 2i\sqrt{2})(-\sqrt{3}- i)$$ $$=4(\text{cos}\hspace{.1em}225° + i \hspace{.1em}\text{sin}\hspace{.1em}225°) \cdot 2(\text{cos}\hspace{.1em}210° + i \hspace{.1em}\text{sin}\hspace{.1em}210°)$$ Let's reference our product theorem: $$[r_{1}(\text{cos}\hspace{.1em}θ_{1}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1})] \cdot [r_{2}(\text{cos}\hspace{.1em}θ_{2}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{2})]=r_{1}\cdot r_{2}[\text{cos}(θ_{1}+ θ_{2}) + i \hspace{.1em}\text{sin}(θ_{1}+ θ_{2})]$$ $$ 4(\text{cos}\hspace{.1em}225° + i \hspace{.1em}\text{sin}\hspace{.1em}225°) \cdot 2(\text{cos}\hspace{.1em}210° + i \hspace{.1em}\text{sin}\hspace{.1em}210°)$$ $$=4 \cdot 2[\text{cos}(225° + 210°) + i \hspace{.1em}\text{sin}(225° + 210°)]$$ $$=8 (\text{cos}\hspace{.1em}435° + i \hspace{.1em}\text{sin}\hspace{.1em}435°)$$ 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 in polar form: $$-\frac{\sqrt{14}}{2}- \frac{\sqrt{14}}{2}i$$ 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: $$\frac{r_{1}(\text{cos}\hspace{.1em}θ_{1}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{1})}{r_{2}(\text{cos}\hspace{.1em}θ_{2}+ i \hspace{.1em}\text{sin}\hspace{.1em}θ_{2})}=\frac{r_{1}}{r_{2}}[\text{cos}(θ_{1}- θ_{2}) + i \hspace{.1em}\text{sin}(θ_{1}- θ_{2})]$$ $$\frac{2(\text{cos}\hspace{.1em}300° + i \hspace{.1em}\text{sin}\hspace{.1em}300°)}{\sqrt{7}(\text{cos}\hspace{.1em}225° + i \hspace{.1em}\text{sin}\hspace{.1em}225°)}$$ $$=\frac{2}{\sqrt{7}}(\text{cos}(300° - 225°) + i \hspace{.1em}\text{sin}(300° - 225°)$$ $$=\frac{2\sqrt{7}}{7}(\text{cos}\hspace{.1em}75° + i \hspace{.1em}\text{sin}\hspace{.1em}75°)$$

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