Integrate $f(z)=e^{3z}$ along
- line segment from $(0,0)\to(1,1)$
- parabola $y=x^2$ from $(0,0)\to(1,1)$
- circle $|z|=3$ once around its arc (positive $360^o$)
First I parametrized with $z(t)=t+it$ and $z'(t)=1+i$ when $t\in[0,1]$ $$\int_0^1 e^{3(t+it)} (1+i)dt=\frac 13 \Bigg|_0^1 e^{3(t+it)}=\frac 13(e^{3+3i}-1)=-\frac13+\frac 13e^3\cos3+i\frac13e^3\sin3 $$
Or do I have to separate the integral with euler to $\int \operatorname{Re}+i\int\operatorname{Im}$ before I integrate?
In seconds $z(t)=t+it^2$ and $z'(t)=1+i2t$ when $t\in[0,1]$ $$\int_0^1 e^{3(t+it^2)}(t+i2t)dt=\frac 13 \Bigg|_0^1 e^{3(t+it^2)}= \text{same as first one it seems...}$$
For the third I think we need to parametritze $x=r\cos t$ and $y=r\sin t$ so $z'(t)=3\cos t+i3\sin t$ and $z'(t)=-3\sin t+ i3 \cos t$ when $t\in[0,2\pi]$
$$\int_0^{2\pi}e^{9(\cos t + i\sin t)}(-3\sin t+ i3 \cos t)dt=\frac 13 \Bigg|_0^{2\pi} e^{9(r\cos t + i\sin t)}=\frac 13(e^{9\cos 2\pi}-e^{9\cos0})=0$$
Does my work make sense?