Question 882013: Rewrite the numerator and denominator in polar or exponential form before carrying out the division.
(sqrt(3)-i)/(2i+2)
I know how to approach this problem but I just can't seem to get the same answer as in the textbook. This is my working:
(sqrt(3)-i) >>> Modulus = 2, Arg = -arctan(1/sqrt(3))= - pi/6
(sqrt(3)-i)= 2(cos(-pi/6) + sin(-pi/6)i)
(2i+2) >>> Modulus = 2sqrt(2), Arg = arctan(1) = pi/4
(2i+2) = 2sqrt(2)(cos(pi/4) + sin(pi/4)i)
(sqrt(3)-i)/(2i+2) = 2/(2sqrt(2))*e^(-pi/6 - pi/4) = 1/sqrt(2) * e^(-5pi/12)
The textbook answer is sqrt(2) * e^(-5pi/12). Who is correct? Thank you.
Answer by pj66(7)   (Show Source):
You can put this solution on YOUR website! I don't really understand the e to the power -pi/6 -pi/4 part, so, I solve in standard form(without e to the power).
All that you have solved is correct. I think the problem is in the division part.
After putting the values, we get 2*((cos pi/6)-(sin pi/6)i)/(2^(3/2)((cos pi/4)+(sin pi/4)i))
where -cos pi/6=cos pi/6
multiply and divide by (cos pi/4)-(sin pi/4)i)
=> ((cos pi/6 - sin pi/4i)*(cos pi/4 - sin pi/4i))/(sqrt2(cos pi/4 + sinpi/4i)*(cos pi/4 - sin pi/4i))
=>((cos pi/6)*(cos pi/4)-(sin pi/6)*(cos pi/4)i-(cos pi/6)*(sin pi/4)i-(sin pi/6)*(sin pi/4))/(sqrt 2)
Using cos(x+y)=cosx*cosy-siny*cosx
and sin(x+y)=sinx*cosy-siny*cosx
=>(2^(-1/2))*(cos(pi/6 + pi/4)-sin(pi/6 + pi/4)i)
I hope this helps.
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