Transforming complex integration to real integraton

Question

Suppose $z=x+iy$ is a complex number, and $x=Re(z)$ and $y=Im(z)$ are real numbers. Is the following statement correct ( function $f(.)$ has no singularity in $[a,b]$)? $$\int_a^bf(z)dz=\int_{Re(a)}^{Re(b)}\int_{Im(a)}^{Im(b)}f(x+iy)dx dy$$

Answer 1

If by $\int_a^bf(z)dz$ you mean the integral of $f(z)$ along the line from $a$ to $b$ along the real line, you need not consider any complex arguments, since the complex variable $z$ is parameterized by a curve whose entire range is real. This would make the integral the ordinary real integral $\int_a^b f(x)dx$ where $a \leq x \leq b$.

To the point of @Kaynex, in the case of complex $a$ and $b$, there will be a parameterization $\gamma(t) = c t + d$ where $c$ and $d$ are complex and $0 \leq t \leq 1$. It follows that $dz = \gamma' dt = cdt$ and therefore $$ \int_a^bf(z)dz = c\int_0^1 f(ct+d)dt. $$