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FP2: Complex numbers problem.

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.

Cheers.
Reply 1
Avatar for ttoby
ttoby
15
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.
Well aren't I a daft ****.

Thanks.
Reply 3
Avatar for DPLSK
DPLSK
14
TheDefiniteArticle
Show that 1+ejx=2ejx2cos(x2)\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)=ejx+e−jx22cos(x)=\frac{e^{jx}+e^{-jx}}{2}

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

It should make life a lot easier.

I hope it helps.

Darren
(edited 12 years ago)

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