9.7 Products and Quotients of Complex Numbers in Polar Form

1. 9.7 Products and Quotients of Complex Numbers in Polar Form The product of two complex numbers, and Can be obtained by using the following formula:

2. Products and Quotients of Complex Numbers in Polar Form The quotient of two complex numbers, and Can be obtained by using the following formula:

3. Products and Quotients of Complex Numbers in Polar Form Find the product of 5cis30 and –2cis120 Next, write that product in rectangular form

4. Products and Quotients of Complex Numbers in Polar Form Find the quotient of 36cis300 divided by 4cis120 Next, write that quotient in rectangular form

5. Products and Quotients of Complex Numbers in Polar Form Find the result of Leave your answer in polar form. Based on how you answered this problem, what generalization can we make about raising a complex number in polar form to a given power?

6. De Moivre’s Theorem De Moivre's Theorem is the theorem which shows us how to take complex numbers to any power easily. De Moivre's Theorem – Let r(cos F+isin F) be a complex number and n be any real number. Then [r(cos F+isin F]n = rn(cos nF+isin nF) What is this saying? The resulting r value will be r to the nth power and the resulting angle will be n times the original angle.

7. De Moivre’s Theorem Try a sample problem: What is [3(cos 45°+isin45)]5? To do this take 3 to the 5th power, then multiply 45 times 5 and plug back into trigonometric form. 35 = 243 and 45 * 5 =225 so the result is 243(cos 225°+isin 225°) Remember to save space you can write it in compact form. 243(cos 225°+isin 225°)=243cis 225°

8. De Moivre’s Theorem Find the result of: Because of the power involved, it would easier to change this complex number into polar form and then use De Moivre’s Theorem.

9. De Moivre’s Theorem De Moivre's Theorem also works not only for integer values of powers, but also rational values (so we can determine roots of complex numbers).

10. De Moivre’s Theorem Simplify the following:

11. Complex Root Theorem Every complex number has ‘p’ distinct ‘pth’ complex roots (2 square roots, 3 cube roots, etc.) To find the p distinct pth roots of a complex number, we use the Complex Root Theorem (which is related to De’Moivre’s Theorem) …where ‘n’ is all integer values between 0 and p-1. Why the 360? Well, if we were to graph the complex roots on a polar graph, we would see that the p roots would be evenly spaced about 360 degrees (360/p would tell us how far apart the roots would be).

12. Complex Root Theorem Find the 4 distinct 4th roots of -3 - 3i

13. Complex Root Theorem Solve the following equation for all complex number solutions (roots):