# Complex Integration with a Modulus

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.

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

Hint: $$|a+bi|^2 = a^2 + b^2$$, where $$a$$ and $$b$$ are real.
• 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)? – Mathstatsstudent Mar 4 '19 at 22:34
• $t$ is a real parameter that varies between $0$ and $1$. Note your limits of integration are also wrong and should therefore be $0$ and $1$. – MPW Mar 5 '19 at 0:21
• $t$ 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. – MPW Mar 5 '19 at 0:25