- #1
futurebird
- 272
- 0
I've been asked to find the value of:
[tex]\int_{-1}^{1}z^{\frac{1}{2}}dz[/tex]
Here is how I did it:
I want to change to polar form. To do that I'll use the fact that:
[tex]z=e^{i\theta}[/tex] For a unit circle radius 1.
[tex]dz=ie^{i\theta}d\theta[/tex]
I replace z and dz:
[tex]= \int_{-1}^{1}e^{\frac{i\theta}{2}}ie^{i\theta}d\theta[/tex]
[tex]= \int_{-1}^{1}ie^{\frac{i\theta}{2}}e^{i\theta}d\theta[/tex]
[tex]= \int_{-1}^{1}ie^{\frac{3i\theta}{2}}d\theta[/tex]
[tex]= \frac{2}{3}\int_{-1}^{1}\frac{3i}{2}e^{\frac{3i\theta}{2}}d\theta[/tex]
[tex]= \frac{2}{3}(e^{\frac{3i\theta}{2}})^{-1}_{1}[/tex]
[tex]= (\frac{2}{3})(e^{\frac{3i}{2}}-e^{\frac{-3i}{2}})[/tex]
Did I do this correctly? How can I check my work?
[tex]\int_{-1}^{1}z^{\frac{1}{2}}dz[/tex]
Here is how I did it:
I want to change to polar form. To do that I'll use the fact that:
[tex]z=e^{i\theta}[/tex] For a unit circle radius 1.
[tex]dz=ie^{i\theta}d\theta[/tex]
I replace z and dz:
[tex]= \int_{-1}^{1}e^{\frac{i\theta}{2}}ie^{i\theta}d\theta[/tex]
[tex]= \int_{-1}^{1}ie^{\frac{i\theta}{2}}e^{i\theta}d\theta[/tex]
[tex]= \int_{-1}^{1}ie^{\frac{3i\theta}{2}}d\theta[/tex]
[tex]= \frac{2}{3}\int_{-1}^{1}\frac{3i}{2}e^{\frac{3i\theta}{2}}d\theta[/tex]
[tex]= \frac{2}{3}(e^{\frac{3i\theta}{2}})^{-1}_{1}[/tex]
[tex]= (\frac{2}{3})(e^{\frac{3i}{2}}-e^{\frac{-3i}{2}})[/tex]
Did I do this correctly? How can I check my work?