My teacher said without explaining that since the period of $\cos(z)$ is $2\pi$ then the period of $\cos(2iz+13)$? is $i\pi$. Why is this true?
Why the period of $\cos(2iz+13)$ is $i\pi$?
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complex-analysis
complex-numbers
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0Try expanding out $\cos(2i(z+i\pi)+13)$. – 2017-01-07
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1Generalize the result to make it easier: if $f$ has period $p$ then $g$ defined by $g(x)=f(ax+b)$ with $a\ne0$ has period $p/a$. Now your proof! – 2017-01-07
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0As an aside, the use of the article 'the' implies uniqueness of a period. This isn't right. In this context there are infinite periods. – 2017-01-07
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0@Git, right; the teacher ought to have spoken of "minimal periods"... – 2017-01-07
1 Answers
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We know that the function $\cos z $ is $2\pi $ periodic. Let the period of $\cos (2iz+13) $ be $k $. Since $$\cos [2i (z+k)+13]=\cos [2iz +13+2ik]$$ it follows that $2ik =2\pi \Rightarrow k=-i\pi $ and hence the required period $k $ is $i\pi $. Hope it helps.