Modulus of a Complex Logarithm

Question

I'm currently self-studying complex analysis and I'd like to analytically show that $$\lim\limits_{R\to\infty}\int_{\gamma_R} \frac{\ln\left(z+i\right)}{z^2+1}\ \mathrm dz=0$$ Where $$ \gamma_R=\left\{Re^{it}:t\in [0, \pi]\right\}$$ After setting $z=Re^{it}$, using the triangle inequality and applying the estimation lemma, I end up with $$\int_0^{\pi} \left|\frac{\ln\left(Re^{it}+i\right)}{\left(Re^{it}\right)^2+1}\right|\left|iRe^{it}\right|\ \mathrm dt\leq\frac{\left|\ln\left(Re^{it}+i\right)\right|}{R^2-1}\int_0^{\pi} \left|iRe^{it}\right|\ \mathrm dt$$ $$\int_0^{\pi} \left|\frac{\ln\left(Re^{it}+i\right)}{\left(Re^{it}\right)^2+1}\right|\left|iRe^{it}\right|\ \mathrm dt\leq\frac{\pi R\left|\ln\left(Re^{it}+i\right)\right|}{R^2-1}$$ I'm intuitively aware that $$\lim\limits_{R\to\infty}\frac{\pi R\left|\ln\left(Re^{it}+i\right)\right|}{R^2-1}=0$$ My question is how do I proceed in analytically showing that the aforementioned limit is indeed zero? My trouble seems to come from a lack of understanding of how the modulus of a complex log works. Any additional insight on this is very appreciated and thank you for reading my post.

Answer 1

To be precise, let's cut the plane at the branch point $z=-i$ with a straight line along the negative imaginary axis to $z=-i\infty$. Then, on $\gamma_R$, we have

$$0<\arctan(1/R)\le \arg(z+i)\le \pi -\arctan(1/R)<\pi$$

Therefore, on $\gamma_R$, the magnitude of the complex logarithm is bounded by

$$\begin{align}|\log(Re^{i\phi}+i)|&\le \sqrt{\log^2\left(\sqrt{R^2+2R\sin(\phi)+1}\right)+\pi^2}\\\\ &\le \sqrt{\log^2\left(R+1\right)+\pi^2} \end{align}$$

Finally, we can bound the integral of interest by

$$\begin{align} \left|\int_{\gamma_R}\frac{\log(z+i)}{z^2+1}\,dz\right|&=\left|\int_0^\pi \frac{\log(Re^{i\phi}+i)}{R^2e^{i2\phi}+1}\,iRe^{i\phi}\,d\phi\right|\\\\ &\le \int_0^\pi \left|\frac{\log(Re^{i\phi}+i)}{R^2e^{i2\phi}+1}\,iRe^{i\phi}\right|\,d\phi\\\\ &\le \int_0^\pi \frac{|\log(Re^{i\phi}+i)|}{|R^2e^{i2\phi}+1|}|iRe^{i\phi}|\,d\phi\\\\ &=\int_0^\pi \frac{|\log(Re^{i\phi}+i)|}{|R^2e^{i2\phi}+1|} R\,d\phi\\\\ &\le \frac{\pi\,R\,\sqrt{\log^2\left(R+1\right)+\pi^2} }{R^2-1}\\\\ &\to 0\,\,\text{as}\,\,R\to \infty \end{align}$$