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Math Help - simplifying complex number to rcisΘ form?

  1. #1
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    simplifying complex number to rcisΘ form?

    Hey guys i'm hoping someone can help me with with problem.
    i found z=-2+2i in rcisΘ (r.Θ) being (2√2, 3π/4)

    I need to find w= √2cis 45 find zw in rcisΘ form and the reciprocal of z in a+bi form.

    so far i got (1+1i) for w but i don't know if that is right.

    Any help is appreciated!
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  2. #2
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    Re: simplifying complex number to rcisΘ form?

    Do you know that if \displaystyle \begin{align*} z_1 = r_1\,\textrm{cis}\,{(\theta _1)} \end{align*} and \displaystyle \begin{align*} z_2 = r_2\,\textrm{cis}\,{(\theta _2)} \end{align*}, then \displaystyle \begin{align*} z_1z_2 = r_1r_2\,\textrm{cis}\,{ \left( \theta_1 + \theta_2 \right) } \end{align*}?
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  3. #3
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    Re: simplifying complex number to rcisΘ form?

    no i did not know that.
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  4. #4
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    Re: simplifying complex number to rcisΘ form?

    would that be like
    √2 x 2√2 (cos (π/4 + 3π/4) + isin (π/4 + 3π/4)
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  5. #5
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    Re: simplifying complex number to rcisΘ form?

    Assuming you meant to close that last bracket, then yes.
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  6. #6
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    Re: simplifying complex number to rcisΘ form?

    Awesome. Thanks so much!
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