Recent content by diddy_kaufen

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    What is the solution to the complex cosine equation without using logarithms?

    Ok I get $$e^x (\cos y + i \sin y) = \frac{1}{2}+p$$ or $$e^x \cos y = \frac{1}{2}+p$$ can I work around this without using the logarithm?
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    What is the solution to the complex cosine equation without using logarithms?

    Thank you for your help, I'm sorry but I'm still stuck. I used the relevant equations to substitute w = e^(iz)to get w=0 or w=-1, so |w|=e^(-y)=0 or |w|=1, so $$e^{iz}=e^{-y}(\cos x_0 + i \sin x_0)$$ which means that $$ x_0=\pi+n2\pi $$ is this even correct? I'm sorry I'm new to complex numbers
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    What is the solution to the complex cosine equation without using logarithms?

    but wouldn't I then still be using the logarithm? Since then I would have to do 1/2+p = exp( ln( 1/2+p ) ) but I'm not getting all solutions, since p is in Z and I'm not allowed to evaluate a complex logarithm?
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    What is the solution to the complex cosine equation without using logarithms?

    Homework Statement Solve the equation $$cos(\pi e^z) = 0$$Homework Equations I am not allowed to use the complex logarithm identities. $$ \cos z = \frac{e^{iz}+e^{-iz}}{2} $$ $$e^{i\theta}=\cos\theta+i \sin\theta$$ The Attempt at a Solution All I've gotten is $$\cos(\pi e^z)=0 \iff \pi...
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    F non-measurable but |f| and f^2 are

    Homework Statement Find a function in a measurable space that is non-measurable, but |f| and f2 are measurable. Homework Equations None. The Attempt at a Solution I am trying to understand the following answer to the problem: (source: http://math.stackexchange.com/a/1233792/413398) I do...
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    Show a limited function is measurable

    Not sure about the translated term limited (from German); perhaps cut-off function? Homework Statement Let f be a measurable function in a measure space (\Omega, \mathcal{F}, \mu) and C>0. Show that the following function is measurable: f_C(x) = \left\{ \begin{array}{ll} f(x) & \mbox{if }...
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