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anonymous

  • one year ago

Find the product of the complex numbers. Express your answer in trigonometric form. z1=3(cos(pi/4)+i sin(pi/4)) z2=7(cos(3pi/8)+i sin(3pi/8))

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  1. tkhunny
    • one year ago
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    Have you considered multiplying the leading coefficients and adding the angles?

  2. anonymous
    • one year ago
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    so 3*7?

  3. tkhunny
    • one year ago
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    Now, the second part.

  4. anonymous
    • one year ago
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    multiply pi/4*pi/4?

  5. tkhunny
    • one year ago
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    Read my first post.

  6. anonymous
    • one year ago
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    Add pi/4+pi/4?

  7. tkhunny
    • one year ago
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    Why would you do that? When you came up with 3*7, you took a value from each of z1 and z2. You should do the same with the addition.

  8. anonymous
    • one year ago
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    you add 3 and 7? is that what you're saying?

  9. tkhunny
    • one year ago
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    No, Please don't reverse correct things you did already. Multiply the leading constants: 3*7 -- Done. Don't change that. Add the angles: ?? + ??

  10. anonymous
    • one year ago
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    ok I think I understand I use the unit circle to figure out the angle then I add?

  11. tkhunny
    • one year ago
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    ?? The angles are given. Why do you need a unit circle? \(\dfrac{\pi}{4}+\dfrac{3\pi}{8}\) -- Go!

  12. anonymous
    • one year ago
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    5pi/8

  13. tkhunny
    • one year ago
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    Now, if this happens to be greater than \(2\pi\), you might need that Unit Circle. What is your final answer? You have all the pieces.

  14. anonymous
    • one year ago
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    So do I do 21cos(5pi/8)+i sin(5pi/8) ?

  15. tkhunny
    • one year ago
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    Not quite. The parentheses are not optional.

  16. anonymous
    • one year ago
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    21(cos(5pi/8)+i sin(5pi/8))

  17. tkhunny
    • one year ago
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    Perfect.

  18. anonymous
    • one year ago
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    okay let me solve

  19. tkhunny
    • one year ago
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    Solve what? You are done. Move on to the next one.

  20. anonymous
    • one year ago
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    oh that's it?

  21. tkhunny
    • one year ago
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    That's all there is to it. :-)

  22. anonymous
    • one year ago
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    thanks

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