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$\gamma(t)=Re^{it}$ for $0\le t \le 2\pi$, $R\in \mathbb{R}$.

In my notes it said the length of $\gamma$ =$\int_0^{2\pi}|\gamma'(t)|dt=2\pi R$.

Intuitively, it makes sense that it is the circumference of the circle, but calculating it out, I get a different answer?

$$\int_0^{2\pi}Rie^{it}dt=Ri[\frac{e^{it}}{i}]_0^{2\pi}\\=R[e^{2\pi i}-e^0]\\=R[e^1-1]?$$

The reason why I'm trying to calculate $\gamma$ =$\int_0^{2\pi}|\gamma'(t)|dt$ is because it is needed for the estimation lemma.

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You want the magnitude of $Re^{it}$. –  Jackson Walters May 24 '12 at 23:02
    
Thanks Jackson, didnt realise it was the magnitude! –  Derrick May 24 '12 at 23:08

1 Answer 1

up vote 4 down vote accepted

Note that the integrand is $\lvert \gamma'(t) \rvert$ and not just $\gamma'(t)$. We have that $\gamma'(t) = iR e^{it}$.

Remember that $\lvert z_1 z_2 z_3 \rvert = \lvert z_1 \rvert \lvert z_2 \rvert \lvert z_3 \rvert $.

Hence, $$\lvert \gamma'(t) \rvert = \lvert i \rvert \lvert R \rvert \lvert e^{i t} \rvert$$

Now $\lvert e^{it} \rvert = 1$ since $\cos^2(t) + \sin^2(t) = 1$.

Also, $\lvert i \rvert = \sqrt{0^2 + 1^2} = 1$.

Hence, $\lvert \gamma'(t) \rvert = 1 \times R \times 1$. (Since $\lvert i \rvert = 1$ and $\lvert e^{it} \rvert = 1$.)

Hence, the length of the curve is $$\int_0^{2 \pi} \lvert \gamma'(t) \rvert dt = \int_0^{2 \pi} R dt = 2\pi R$$ which is precisely what you expect and want.

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Thanks @Marvis , I understand what you did, except, how did you get that $\lvert \gamma'(t) \rvert = R$? I know that $z=x+iy$, then $|z| = \sqrt {x^2+y^2}$, but what formula//what exactly did you do to get $\lvert \gamma'(t) \rvert = R$? –  Derrick May 24 '12 at 23:08
1  
@Derrick I have updated the answer. Hope it clarifies your question. –  user17762 May 24 '12 at 23:15
    
Thank you again, understood it fully now. :) –  Derrick May 24 '12 at 23:16

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