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Write z = 2(sqrt)2 + 2(sqrt)2i in polar form.

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

Write z = -6i in polar form.

z = 6 cis 3pi/2
z = 6 cis pi/2
z = -6 cis 3pi/2
x = 6(sqrt)2 cis 3pi/2

I've tried figuring these two out but I need help

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4 answers
  1. z = 2√2 + 2√2 i
    r = √(8+8) = 4
    tanØ = 2√2/(2√2) = 1
    Ø = π/4

    z = 4(cosπ/4 + i sinπ/4) or 4cisπ/4 , if you learned that abbreviation

    for z = -6i
    consider it as z = 0 - 6i and proceed as before
    from the graph in the Argand plane , it can be seen that the angle is 3π/2

    which of the answer choices would apply ?

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  2. The third one: z = -6 cis 3pi/2

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  3. Whoops I'm wrong it's the first one

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  4. Yup, it is +6
    the r value is considered positive, the direction will be take care
    of by the angle.

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