Complex Integrals (No Residue allowed)

Question

Complete the integrals along curve $C$

a) $\displaystyle\int_C\frac1z\ \mathrm dz$

b) $\displaystyle\int_Cf(z)\ \mathrm dz;\quad[f(a)]^2=z\ \&\ f(1)=1$

c) $\displaystyle\int_Cg(z)\ \mathrm dz;\quad g(z)=\sqrt[4]z\ \&\ g(1)=i$

Answer 1

a) Close the curve by adding the segment $[1,e]$ with the proper orientation to it. I will call the new curve $c^{+}$. Use the definition of index of the curve in the origin:

$2\pi i Ind_{c^{+}}(0)=\int_{c^{+}}\frac{1}{z} dz=\int_{c}\frac{1}{z} dz+\int_{[1,e]}\frac{1}{z} dz=\int_{c}\frac{1}{z} dz+(Log(1)-Log(e))=\int_{c}\frac{1}{z} dz-1$

Hence $\int_{c}\frac{1}{z} dz=4\pi i +1$

b) $f(z)=\sqrt z$ or $f(z)=-\sqrt z$ but since $f(1)=1$, we deduce $f(z)=\sqrt z$. A primitive of $f(z)$ is $\frac{2}{3}z^{\frac{3}{2}}$ in $\mathbb{C}$.

Hence, $\int_{c}f(z)dz=\frac{2}{3}(e^{\frac{3}{2}}-1)$

c) $g(z)$, defined as you did, is not the function you are looking for, since $g(1)$ would be $1$. So I am assuming that our function $g$ is one of the fourth roots of $z$, in particular the one verifying the condition $g(1)=i$. This is the case for $g(z)=|z|^{\frac{1}{4}}(cos(\frac{arg(z)+2\pi}{4})+isin(\frac{arg(z)+2\pi}{4}))$, i.e. $g(z)=iz^{\frac{1}{4}}$. A primitive of $g(z)$ is $\frac{4i}{5}z^{\frac{5}{4}}$ in $\mathbb{C}$.

Hence, $\int_{c}g(z)dz=\frac{4i}{5}(e^{\frac{5}{4}}-i^{\frac{5}{4}})$