I need to find $M_0$ and $k>1$ constants such that $|\frac{\ln(z+i\sqrt{5})}{z^{2} +7}|\leq \frac{M_{0}}{R^{k}}$ with $z=Re^{i\theta}$.
This is what I got so far:
Since $|z|^{2} -7 \leq |z^{2}+7|$, then $$\frac{1}{|z^{2} +7|}\leq\frac{1}{|z^{2}|-7}$$
Thus multiplying both sides $$\Bigg|\frac{\ln(z+i\sqrt{5})}{z^{2}+7}\Bigg|\leq\frac{|\ln(z+i\sqrt{5})|}{|z|^{2}-7}$$
Since $|z|^{2}=|Re^{i\theta}|^{2}=R^{2}$
$$\Bigg|\frac{\ln(z+i\sqrt{5})}{z^{2}+7}\Bigg| \leq \frac{|\ln(z+i\sqrt{5})|}{R^{2}-7}$$
What should I do for this $|\ln(z+i\sqrt{5})|$ to become a constant?
You're nearly there. Now, by choosing the principal branch for $\ln z$ since it's a multi-valued function. You would have: $\log z = \log|z| + i\arg z$. Hence your inequality of the last part would be: $$\frac{|\ln(z+i\sqrt{5})|}{R^{2}-7}= \frac{\left|\ln |z+i\sqrt{5}|+i\arg (z+i\sqrt{5})\right |}{R^{2}-7}\leq \frac{\ln |z+i\sqrt{5}|+\pi}{R^{2}-7}$$$$\leq \frac{ |z+i\sqrt{5}|+\pi}{R^{2}-7}\leq\frac{ |z|+|i\sqrt{5}|+\pi}{R^{2}-7}=\frac{ R+\sqrt{5}+\pi}{R^{2}-7} $$