This is an exercise problem, which asks to prove the non-existence of continuous logairthm on $\mathbb{C}^*$, Using the results of path integrals, In that book they already proved the non-existence using the non-existence of continuous argument.
I tried to give a proof; I want someone to verify it
suppose there exists a continuous logarithm on $\mathbb{C}^*$, then by derivatives of inverse function theorem $\log^{'}(z)=\frac{1}{z}$, Now consider the path integral $\int_{\gamma}\frac{1}{z}$,where $\gamma:[0,2\pi]\to \mathbb{C}$, such that $\gamma(t)=e^{it}$
As the primitive of $\frac{1}{z}$ exists, by the fundamental theorem of path integrals, $$\int_{\gamma}\frac{1}{z}=\log(\gamma(1))-\log(\gamma(0))=0$$
But, we know that $$\int_\gamma\frac{1}{z}=\int_{0}^{2\pi}\frac{1}{e^{it}}.ie^{it} dt$$ $$=2\pi i$$
And hence a contradiction. Therefore there exists no continuous logarithm on $\mathbb{C}^*$.
Is this proof looks neat and fine, If not what's wrong, Kindly help. Thanks in Advance