FP2: Complex numbers problem.

    • Thread Starter

    Hi, I've been set a homework sheet and think I've made a mistake somewhere along this question:

    Show that 1 + e^(jx) = (2cos((1/2)x))e^((1/2)jx)

    My working thus far (may have mistakes in, I'm tired and braindead):

    1 + e^(jx)
    = 1 + cosx + jsinx
    = 1 + (1/2)(2cos^2((1/2)x) + 1) + 2jsin((1/2)x)cos((1/2)x)
    = (3/2) + cos^2((1/2)x) + 2jsin((1/2)x)cos((1/2)x)
    = (3/2) + (2cos((1/2)x)(1/2cos((1/2)x) + jsin((1/2)x))

    And naturally I would turn that last bit into e^((1/2)jx) if I had a whole cos((1/2)x), but it still wouldn't be the result I want.

    Sorry if this is hard to read, but I'm hopeless with LaTeX.


    On this line:

    (Original post by TheDefiniteArticle)
    = 1 + (1/2)(2cos^2((1/2)x) + 1) + 2jsin((1/2)x)cos((1/2)x)
    The 1/2 shouldn't be there, and the +1 should be -1.

    You should find that things cancel/simplify from here.
    • Thread Starter

    Well aren't I a daft ****.


    (Original post by TheDefiniteArticle)
    Show that \displaystyle 1+e^{jx}=2e^{\frac{jx}{2}} cos \left( \frac{x}{2} \right).
    I personally would have started with the RHS, using the following identity. 2cos(x)=\frac{e^{jx}+e^{-jx}}{2}

    Aside: This can be shown using the fact that e^{jx}=cos(x)+jsin(x).

    It should make life a lot easier.

    I hope it helps.

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Updated: March 24, 2012
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