[3/4(cos pi/3 + i sin pi/3)][4(cos 3pi/4 + i sin 3pi/4)]
= (3/4)(1/2 + i (√3/2) (4) (-√2/2 + i √2/2)
= (3/4)(1/2)(1 + i √3)(4)(1/2)( -√2 + i √2)
= (3/4)(-√2 + i√2 - i √6 + √6 i^2)
= (3/4) (-√2 -√6 + i(√2-√6) ) --- in quad III
tan^-1 | √2-√6|/|-√2-√6| = 15° or π/12
but the angle is in III , so Ø = 135° or 9π/12
= (3/4)( cos 9π/12 + i sin 9π/12)
check my arithmetic, I should have written it out on paper first.
1) Perform the operation and leave the result in trig. form.
[3/4(cos pi/3 + i sin pi/3)][4(cos 3pi/4 + i sin 3pi/4)]
Thanks
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