How do you simplify #2(cos((3pi)/4)+isin((3pi)/4))*sqrt2(cos(pi/2)+isin(pi/2))# and express the result in rectangular form?

1 Answer
Jan 11, 2017

Answer:

#2(cos((3pi)/4)+isin((3pi)/4))*sqrt2(cos(pi/2)+isin(pi/2))=-2-2i#

Explanation:

#2(cos((3pi)/4)+isin((3pi)/4))*sqrt2(cos(pi/2)+isin(pi/2))#

= #2sqrt2{cos((3pi)/4)cos(pi/2)+icos((3pi)/4)sin(pi/2)+icos(pi/2)sin((3pi)/4)+i^2sin((3pi)/4)sin(pi/2)}#

= #2sqrt2{cos((3pi)/4)cos(pi/2)-sin((3pi)/4)sin(pi/2)+i(cos((3pi)/4)sin(pi/2)+icos(pi/2)sin((3pi)/4))}#

= #2sqrt2{cos((3pi)/4+pi/2)+isin((3pi)/4+pi/2)}#

= #2sqrt2(cos((5pi)/4)+isin((5pi)/4))#

= #2sqrt2(-cos(pi/4)-isin(pi/4))#

= #2sqrt2(-1/sqrt2-i1/sqrt2)#

= #-2-2i#