How do you multiply: 2(cos(pi/3)+ i sin(pi/3)) and 4(cos(pi/4)+ i sin(pi/4))?

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Answer 1 · 2016-06-24T04:59:51

Product is $8e^(i(7pi)/12)=8(cos((7pi)/12)+isin((7pi)/12))$

= $(2sqrt2-2sqrt6)+i(2sqrt6+2sqrt2)$

To multiply the two complex numbers in polar form there could be three ways.

A - $(2(cos(pi/3)+ i sin(pi/3)))xx(4(cos(pi/4)+ i sin(pi/4)))$

= $8(1/2+ i sqrt3/2)(1/sqrt2+ i1/sqrt2)$

= $8(1/(2sqrt2)+isqrt3/(2sqrt2)+i1/(2sqrt2)-sqrt3/(2sqrt2))$

= $(2sqrt2-2sqrt6)+i(2sqrt6+2sqrt2)$

B - As $(2(cos(pi/3)+ i sin(pi/3)))=2e^(i(pi/3))$ and

$4(cos(pi/4)+ i sin(pi/4))=4e^(i(pi/4))$ , their product is

$4xx2xxe^(i(pi/3+pi/4))=8e^(i(7pi)/12)$

C - $(2(cos(pi/3)+ i sin(pi/3)))xx(4(cos(pi/4)+ i sin(pi/4)))$

= $8[(cos(pi/3)cos(pi/4)-sin(pi/3)sin(pi/4))+i(sin(pi/3)cos(pi/4)+cos(pi/3)sin(pi/4))]$

= $8cos((pi/3)+(pi/4))+isin((pi/3)+(pi/4))$

= $8(cos((7pi)/12)+isin((7pi)/12))$