SOLUTION: Sir, my question is how to dissolve imaginary part in an equation completely. the equation is of the form cos(theeta)+jsin(theeta) & j,the imaginary part should be completely ta

Question 475313: Sir,
my question is how to dissolve imaginary part in an equation completely.
the equation is of the form cos(theeta)+jsin(theeta) & j,the imaginary part should be completely taken out by simplifying & iam not finding the solution.Pls help me
Found 3 solutions by richard1234, ccs2011, robertb:
Answer by richard1234(7193)   (Show Source): You can put this solution on YOUR website!
(more common to use i instead of j for complex numbers...save j for quaternions).

The imaginary part of a complex number in the form a + bi, a & b reals, is just b (not bi). Here, the imaginary part is sin(theta).

Note that this expression is equal to , derived by using Taylor series.
Answer by ccs2011(207)   (Show Source): You can put this solution on YOUR website!
Ok so the equation is a complex number in polar form:

where r is the modulus or the magnitude of the distance from origin on complex plane.
To eliminate the imaginary part, multiply by the conjugate.
The conjugate is just the same number with the imaginary part having the opposite sign.
Ex: z = 2 + 3j, conjugate = 2 - 3j
In polar form the conjugate would be
So multiply by the conjugate:

=
Middle term cancels
=
j^2 = -1
=
trig identity, sin^2 + cos^2 = 1
=
Hope that answered your question.
Answer by robertb(5830)   (Show Source): You can put this solution on YOUR website!
Not exactly sure how your equation looks like, but if it is of the form
,
where is some expression in terms of , then you can use de Moivre's formula to "suppress" j by raising both sides to some integer power n. The only caveat to this is, you have to make sure that
, for some integer k.
If cannot be written in this form then you cannot use de Moivre's theorem to "dissolve" j.
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