
The product z⋅w, where z=2(cos17π36+i17π36) and w=3(cos11π90+isin11π90), and express the answer in polar form and exponential form.

Answer to Problem 10CT
Solution:
The product z⋅w in the polar form is 6(cos107π180+isin107π180) and the exponential form is 6ei⋅107π180.
Explanation of Solution
Given information:
z=2(cos17π36+i17π36) and w=3(cos11π90+isin11π90)
Explanation:
Let z=2(cos17π36+i17π36) and w=3(cos11π90+isin11π90).
z⋅w=2(cos17π36+i17π36)⋅3(cos11π90+isin11π90)
First express the
z⋅w=2ei⋅17π36⋅3ei⋅11π90
The product of z1=r1ei θ1 and z2=r2ei θ2 is find as z1⋅z2=r1r2 ei (θ1+θ2)
⇒z⋅w=2⋅3 ei⋅ (17π36+11π90)
Thus z⋅w=6 ei⋅ 107π180
In the polar form, z⋅w=6(cos107π180+isin107π180).
Therefore, the product z⋅w in polar formis 6(cos107π180+isin107π180), and the exponential form is 6ei⋅107π180.
Chapter 9 Solutions
Precalculus
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