Problem 110 - Algebra and Trigonometry

Question

Prove the rule for finding the quotient of two complex numbers in polar form. Begin the proof as follows, using the conjugate of the denominator's second factor: $$\frac{r_{1}\left(\cos \theta_{1}+i \sin \theta_{1}\right)}{r_{2}\left(\cos \theta_{2}+i \sin \theta_{2}\right)}=\frac{r_{1}\left(\cos \theta_{1}+i \sin \theta_{1}\right)}{r_{2}\left(\cos \theta_{2}+i \sin \theta_{2}\right)} \cdot \frac{\left(\cos \theta_{2}-i \sin \theta_{2}\right)}{\left(\cos \theta_{2}-i \sin \theta_{2}\right)}$$ Perform the indicated multiplications. Then use the difference formulas for sine and cosine.

Step-by-Step Solution

Answer

The rule for finding the quotient of two complex numbers in polar form \(\frac{r_{1}\left(\cos \theta_{1}+i \sin \theta_{1}\right)}{r_{2}\left(\cos \theta_{2}+i \sin \theta_{2}\right)}\) simplifies to \(\frac{r_{1}}{r_{2}} \left(\cos(\theta_{1}- \theta_{2}) + i \sin(\theta_{1}- \theta_{2})\right)\) upon multiplication with the complex conjugate and using difference formulas for sine and cosine.

Step 1: Multiplication with complex conjugate

Multiply the original fraction by \(\frac{\left(\cos \theta_{2}-i \sin \theta_{2}\right)}{\left(\cos \theta_{2}-i \sin \theta_{2}\right)}\). The numerator becomes \(r_{1}r_{2} \left[(\cos \theta_{1} \cos \theta_{2} - i \sin \theta_{1} \cos \theta_{2}) + (i \cos \theta_{1} \sin \theta_{2}-\sin \theta_{1}\sin \theta_{2})\right]\) while the denominator becomes \(r_{2}^2 \left(\cos^2 \theta_{2} + \sin^2 \theta_{2}\right)\) which simplifies to \(r_{2}^2\).

Step 2: Apply difference formulas for sine and cosine

The numerator becomes \(r_{1}r_{2} \left[\cos(\theta_{1}- \theta_{2}) + i \sin(\theta_{1}- \theta_{2})\right]\).

Step 3: Simplify

Finally, the entire expression simplifies to \(\frac{r_{1}}{r_{2}} \left(\cos(\theta_{1}- \theta_{2}) + i \sin(\theta_{1}- \theta_{2})\right)\). This proves the rule for finding the quotient of two complex numbers in polar form.