Let z =38 (cos(pi/8)+ i sin(pi/8)) and w = 2 (cos( pi/16 ) + i sin(pi/16)) What is the product of zw?

Question

Let z =38 (cos(pi/8)+ i sin(pi/8)) and w = 2 (cos( pi/16 ) + i sin(pi/16))

What is the product of zw?

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Gia Bảo 4 years 2021-08-15T16:44:31+00:00 1 Answers 30 views 0

Answers ( )

  1. Philomena
    0
    2021-08-15T16:45:58+00:00
    Reply

    Answer:

    zw=76 * (cos(\frac{3\pi}{16}) +isin(\frac{3\pi}{16}) )

    Step-by-step explanation:

    Given

    z =38 (cos(\frac{\pi}{8})+ i sin(\frac{\pi}{8}) and

    w = 2 (cos(\frac{\pi}{16} ) + i sin(\frac{\pi}{16})

    Required

    Determine zw

    Given that:

    x = v_1(cos(a) + isin(a)) and y = v_2(cos(b) + isin(b))

    x * y = (v_1 * v_2)*((cos(a+b) + isin(a+b))



    So, we have:

    zw=(38 * 2) * (cos(\frac{\pi}{8} + \frac{\pi}{16}) + isin(\frac{\pi}{8} + \frac{\pi}{16}))

    zw=76 * (cos(\frac{2\pi + \pi}{16})+isin(\frac{2\pi + \pi}{16}))

    zw=76 * (cos(\frac{3\pi}{16}) +isin(\frac{3\pi}{16}) )

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