Why no modulus for Complex Integration $\int \frac{1}{z}\:dz$

Question

I am evaluating the integral: $$\int_{\gamma}\frac{dz}{z}$$ Where $\gamma$ is a Straight line path from $1$ to $1+i$.

So by Fundamental theorem of Complex integration we have:

$$\int_{\gamma}\frac{dz}{z}=\int_{1}^{1+i}\frac{dz}{z}=\ln|z|\:\: \big \vert_{1}^{1+i}=\frac{1}{2}\ln(2)$$

But the answer in the book is $$\frac{1}{2}\ln(2)+i\frac{\pi}{4}$$

Answer 1

I suppose the antiderivative of $\frac{1}{z}$ should be $ \operatorname{Log} z$ instead, where $ \operatorname{Log} z$ is the principal branch of the complex logarithm. Accordingly, the value of $\operatorname{Log} (1+i)$ is $ \frac{1}{2} \ln (2) + i \frac{\pi}{4}$.

Answer 2

Let $\gamma\colon[0,1]\longrightarrow\mathbb C$ be defined byt $\gamma(t)=1+ti$. Then that integral is equal to\begin{align}\int_\gamma\frac{\mathrm dz}z&=\int_0^1\frac i{1+ti}\,\mathrm dt\\&=\int_0^1\frac{i+t}{1+t^2}\,\mathrm dt\\&=\int_0^1\frac t{1+t^2}\,\mathrm dt+i\int_0^1\frac1{1+t^2}\,\mathrm dt\\&=\frac12\log(2)+\frac{\pi i}4.\end{align}