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Math Help - Find (Z1)(Z2) Leave answer in polar form

  1. #1
    Junior Member Cyberman86's Avatar
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    Question Find (Z1)(Z2) Leave answer in polar form

    Given:
    Z1= sqrt.6(cos(5pi/6) + i sin(5pi/6))

    Z2= sqrt.2(cos(5pi/4) + i sin (5pi/4))

    Find (Z1)(Z2) Leave answer in polar form

    This is what i got so far and i don't know if i started right.

    =sqrt.12(cos(5pi/6) + i sin(5pie/6)) (cos(5pi/4) + i sin(5pi/4))
    =sqrt.12(cos(5pi/4) + cos(5pie/6) i sin(5pi/4) + i sin(5pi/6) cos(5pi/4) - i sin(5pi/6) sin(5pi/4))

    and i don't know if its right and don't know how to finish it, and if you can, can you show step by step if possible.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Cyberman86 View Post
    Given:
    Z1= sqrt.6(cos(5pi/6) + i sin(5pi/6))

    Z2= sqrt.2(cos(5pi/4) + i sin (5pi/4))

    Find (Z1)(Z2) Leave answer in polar form

    This is what i got so far and i don't know if i started right.

    =sqrt.12(cos(5pi/6) + i sin(5pie/6)) (cos(5pi/4) + i sin(5pi/4))
    =sqrt.12(cos(5pi/4) + cos(5pie/6) i sin(5pi/4) + i sin(5pi/6) cos(5pi/4) - i sin(5pi/6) sin(5pi/4))

    and i don't know if its right and don't know how to finish it, and if you can, can you show step by step if possible.
    Z1= sqrt.6(cos(5pi/6) + i sin(5pi/6))

    Z2= sqrt.2(cos(5pi/4) + i sin (5pi/4))

    z_1=\sqrt{6}e^{\frac{5 \pi}{6}i}

    z_2=\sqrt{2}e^{\frac{5 \pi}{4}i}

    So we get

    z_1 \cdot z_2=\sqrt{6}e^{\frac{5 \pi}{6}i} \cdot \sqrt{2}e^{\frac{5 \pi}{4}i}=2\sqrt{3}e^{\frac{25 \pi}{12}i}=2\sqrt{3}e^{\frac{\pi}{12}i}=2\sqrt{3}\  left[ \cos\left( \frac{\pi}{12}\right)+i\sin\left( \frac{\pi}{12}\right)\right]
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  3. #3
    MHF Contributor red_dog's Avatar
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    If z_1=r_1(\cos\phi_1+i\sin\phi_1) and z_1=r_2(\cos\phi_2+i\sin\phi_2) then

    z_1z_2=r_1r_2[\cos(\phi_1+\phi_2)+i\sin(\phi_1+\phi_2)]
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  4. #4
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    Hello, Cyberman86!

    Given: . \begin{array}{ccc}z_1&=& \sqrt{6}\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right)  \\ \\[-3mm]z_2 &=& \sqrt{2}\left(\cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4}\right) \end{array}

    Find: (z_1)(z_2). .Leave answer in polar form.
    We are expected to know this theorem . . .

    If . \begin{array}{ccc}z_1 &=& r_1\left(\cos\theta_1 + i\sin\theta_2\right) \\ z_2 &=& r_2\left(\cos\theta_2 + i\sin\theta_2\right) \end{array}
    . . . then: . (z_1)(z_2) \;=\;(r_1\cdot r_2)\bigg[\cos\left(\theta_1+\theta_2\right) + i\sin\left(\theta_1+\theta_2\right)\bigg]
    . . . . . . . . . . . . . . . . . . . . . . . . . \uparrow . . . . . . . . \uparrow
    . . . . . . . . . . . . . . . . . . . . . . . . .
    Add the angles!


    We have: . (z_1)(z_2) \;=\;(\sqrt{6}\cdot\sqrt{2})\bigg[\cos\left(\frac{5\pi}{6} + \frac{5\pi}{4}\right) + i\sin\left(\frac{5\pi}{6} + \frac{5\pi}{4}\right) \bigg]

    . . . . . . . . . . . . . = \;2\sqrt{3}\,\left(\cos\frac{25\pi}{12} + i\sin\frac{25\pi}{12}\right)

    . . . . . . . . . . . . . =\;2\sqrt{3}\,\left(\cos\frac{\pi}{12} + i\sin\frac{\pi}{12}\right)



    Edit: Ah, red_dog beat me to it ...
    .
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