Hi having trouble with this question!

Question: Compute line integral -

$\int_\gamma \vert z \vert^2dz$

where $\gamma$ is the line segment from $2$ to $3+i$

Attempt: $\gamma(t) = (1-t)2 + t(3+i)= 2 + 2t + it$

$\gamma'(t) = 2 + i$

$\int_2^{3+i} \vert2 + 2t + it\vert^2 (2+i)dt$

Not sure how to deal with the absolute value or I guess modulus and move forward.

EDIT: I made a mistake when simplifying the parametrization, it's actually

$\int_0^{1} \vert2 + t + it\vert^2 (1+i)dt$

After fixing that and with MPW's comment realized the real part is $(2+t)$ and imaginary just $t$, integrating became easy.

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    $\begingroup$ Your $\gamma(t)$ should be $2 +t +it$, you simplified wrong. So your $\gamma'$ is also wrong. $\endgroup$ – MPW Mar 4 '19 at 21:59
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    $\begingroup$ The limits of integration are also incorrect; $t$ varies from $0$ to $1$. $\endgroup$ – FredH Mar 4 '19 at 22:06

Hint: $|a+bi|^2 = a^2 + b^2$, where $a$ and $b$ are real.

  • $\begingroup$ So using MPW's catch that I simplified wrong; $\int_0^{1} \vert2 + t + ti\vert^2 (1+i)dt$ What part is the real and what part is the imaginary? Getting a little bit confused with the t. Is the real part (2+t) and imaginary just (t)? $\endgroup$ – Mathstatsstudent Mar 4 '19 at 22:34
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    $\begingroup$ $t$ is a real parameter that varies between $0$ and $1$. Note your limits of integration are also wrong and should therefore be $0$ and $1$. $\endgroup$ – MPW Mar 5 '19 at 0:21
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    $\begingroup$ $t$ could appear in either part. $\gamma(t)=(2+t)+ (t)i$, so the real part is $2+t$ and the imaginary part is $t$. The trick there is to separate terms that contain $i$ from terms that don’t. Since $t$ is a real parameter, that makes it easier to see. $\endgroup$ – MPW Mar 5 '19 at 0:25
  • $\begingroup$ Ended up getting the answer after. Thank you! $\endgroup$ – Mathstatsstudent Mar 5 '19 at 5:13

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