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Math Help - Complex No question 1

  1. #1
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    Complex No question 1

    Hi I need help for the following question,

    1.) Let Z1 = 2cis(pi/6) and z2 = rcis(delta) where r>0, and 0<=delta<2pi.
    FInd the range of values of r and delta for which z1z2 is
    a.) a real number greater than 5
    For the modulus of z1z2, I got r>5/2, which was right.
    However for the delta, I'm not sure how I'm supposed to do it, it is (pi/6) + delta >5 and then solve the inequality? Seems wrong..
    b.) a purely imaginary number with modulus less than 1
    How do I find the imaginary number?

    Thank you so much for your time, I really appreciate it!
    J.
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  2. #2
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    Re: Complex No question 1

    Quote Originally Posted by Tutu View Post
    1.) Let Z1 = 2cis(pi/6) and z2 = rcis(delta) where r>0, and 0<=delta<2pi. FInd the range of values of r and delta for which z1z2 is
    a.) a real number greater than 5
    For the modulus of z1z2, I got r>5/2, which was right.
    However for the delta,
    b.) a purely imaginary number with modulus less than 1
    How do I find the imaginary number?
    For z_1z_2 to be a real number greater than 5 you must have
    \frac{\pi}{6}+\delta=2k\pi.

    But also 0\le\delta\le 2\pi.

    b) For z_1z_2 to be purely imaginary number you must have
    \frac{\pi}{6}+\delta be an odd multiple of \frac{\pi}{2}.
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  3. #3
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    Re: Complex No question 1

    Thanks! May I know how you arrived at pi/6 + thetha = 2kpi?

    Why 2kpi, to be exact.

    Thank you so much!
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  4. #4
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    Re: Complex No question 1

    Quote Originally Posted by Tutu View Post
    Thanks! May I know how you arrived at pi/6 + thetha = 2kpi? Why 2kpi, to be exact.

    Suppose that z=r\exp(i\theta) where r\ge 0 is a positive real number if and only if \theta=2k\pi,~k\in\marhbb{Z}.
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  5. #5
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    Re: Complex No question 1

    Is this a rule, thank you!
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    Re: Complex No question 1

    Quote Originally Posted by Tutu View Post
    Is this a rule, thank you!
    I don't know what you mean by rule.

    If z=a+bi and z\in\mathbb{R} it is necessary that b=0.

    If r\cos(\theta)+ir\sin(\theta)\in\mathbb{R} it is necessary that \cos(\theta)=0.
    Thus \theta=k\pi, k\in\mathbb{Z}.

    But in the question, we want a positive real number.
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  7. #7
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    Re: Complex No question 1

    But why is it that when you want a positive real number, it has to be 2kpi?
    I think I kind of understand..
    Does that meant that the angle has to be in the positive direction to get a positive real number hence 2kpi, not -2kpi or anything?
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  8. #8
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    Re: Complex No question 1

    Quote Originally Posted by Tutu View Post
    But why is it that when you want a positive real number, it has to be 2kpi?
    I think I kind of understand..
    Does that meant that the angle has to be in the positive direction to get a positive real number hence 2kpi, not -2kpi or anything?
    \cos([2k+1]\pi)=-1 so it must be an even multiple if \pi.

    So in \theta=2k\pi, the integer k can be positive or negative.

    Recall \cos(\theta)=\cos(-\theta).

    Don't forget that you need 0\le\delta\le 2\pi.
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  9. #9
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    Re: Complex No question 1

    How did you get
    cos((2k+1)pi) = -1?

    More specifically, why 2K+1 and how did you know the whole eqn equals to -1?

    Thank you very much once again!
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