Solve. \large \int_C \pi e^{\pi \overline{z}} \ dz where C is the path that goes from 0 to 1.
logistic_guy Senior Member Joined Apr 17, 2024 Messages 1,610 Apr 4, 2025 #1 Solve. ∫Cπeπz‾ dz\displaystyle \large \int_C \pi e^{\pi \overline{z}} \ dz∫Cπeπz dz where C\displaystyle CC is the path that goes from 0\displaystyle 00 to 1\displaystyle 11.
Solve. ∫Cπeπz‾ dz\displaystyle \large \int_C \pi e^{\pi \overline{z}} \ dz∫Cπeπz dz where C\displaystyle CC is the path that goes from 0\displaystyle 00 to 1\displaystyle 11.
logistic_guy Senior Member Joined Apr 17, 2024 Messages 1,610 Apr 4, 2025 #2 We know that z\displaystyle zz is a complex variable, so it has the form: z=x+iy\displaystyle z = x + iyz=x+iy z‾\displaystyle \overline{z}z is the conjugate of z\displaystyle zz and it has the form: z‾=x−iy\displaystyle \overline{z} = x - iyz=x−iy Since the path lies on the x\displaystyle xx-axis, y=0\displaystyle y = 0y=0 during the whole path. Therefore, z‾=x−i(0)=x\displaystyle \overline{z} = x - i(0) = xz=x−i(0)=x Then, our integral becomes: ∫01πeπx dx\displaystyle \large \int_{0}^{1} \pi e^{\pi x} \ dx∫01πeπx dx
We know that z\displaystyle zz is a complex variable, so it has the form: z=x+iy\displaystyle z = x + iyz=x+iy z‾\displaystyle \overline{z}z is the conjugate of z\displaystyle zz and it has the form: z‾=x−iy\displaystyle \overline{z} = x - iyz=x−iy Since the path lies on the x\displaystyle xx-axis, y=0\displaystyle y = 0y=0 during the whole path. Therefore, z‾=x−i(0)=x\displaystyle \overline{z} = x - i(0) = xz=x−i(0)=x Then, our integral becomes: ∫01πeπx dx\displaystyle \large \int_{0}^{1} \pi e^{\pi x} \ dx∫01πeπx dx
logistic_guy Senior Member Joined Apr 17, 2024 Messages 1,610 Apr 4, 2025 #3 logistic_guy said: ∫01πeπx dx\displaystyle \large \int_{0}^{1} \pi e^{\pi x} \ dx∫01πeπx dx Click to expand... It's straightforward to solve this integral. ∫01πeπx dx=eπx∣01=eπ−1\displaystyle \large \int_{0}^{1} \pi e^{\pi x} \ dx = e^{\pi x}\bigg|_{0}^{1} = e^{\pi} - 1∫01πeπx dx=eπx∣∣∣∣∣01=eπ−1
logistic_guy said: ∫01πeπx dx\displaystyle \large \int_{0}^{1} \pi e^{\pi x} \ dx∫01πeπx dx Click to expand... It's straightforward to solve this integral. ∫01πeπx dx=eπx∣01=eπ−1\displaystyle \large \int_{0}^{1} \pi e^{\pi x} \ dx = e^{\pi x}\bigg|_{0}^{1} = e^{\pi} - 1∫01πeπx dx=eπx∣∣∣∣∣01=eπ−1