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Mathematics 19 Online
OpenStudy (anonymous):

Prove that if z ∈ C and Re(z^k) ≥ 0 for all k = 1, 2, 3, . . ., then z ∈ [0,∞).

OpenStudy (anonymous):

@hartnn Are you able to help?

OpenStudy (zarkon):

use the polar form of a complex number

OpenStudy (anonymous):

so \(z^k=r^k(cos(k\theta) +isin(k\theta)\) and thus \(Re(z^k)=r^kcos(k\theta) \ge 0 \). but i am not entirely sure what the question is asking, specifically the interval given. What actually needs to be proved?

OpenStudy (zarkon):

prove that that \(\theta=0\)

OpenStudy (anonymous):

oh ok that makes more sense to me now as the sine term will vanish. How would i go about proving \(\theta = 0 \)

OpenStudy (zarkon):

Assume \(\theta\ne0\) and derive a contadiction

OpenStudy (zarkon):

you will also have to take care of the times when \(cos(\theta)=0\) ie \(\theta=\pm\dfrac{\pi}{2}\) though those are easy to resolve.

OpenStudy (anonymous):

can you help me with the proof, im not sure how to do it

OpenStudy (zarkon):

show there is some k such that \(\cos(k\theta)<0\)

OpenStudy (anonymous):

i'm stuck i dont know how to start the proof

OpenStudy (anonymous):

@phi you able to help?

OpenStudy (anonymous):

@thomaster Are you able to help?

OpenStudy (anonymous):

@goformit100

OpenStudy (goformit100):

Ya say?

OpenStudy (anonymous):

are you able to help me with this proof?

OpenStudy (goformit100):

Sorry Sir I don't know it's proof

OpenStudy (anonymous):

ok no worries

OpenStudy (zarkon):

if i give you much more...i might as well do the entire problem

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