- #1
John O' Meara
- 325
- 0
Evaluate \int_C f(z)dz by theorem 1 and check your result by theorem 2 where f(z) = z^4 and C is the semicircle |z|=2 from -2i to 2i in the right half-plane.
Theorem 1 : \int_{z_0} ^{z_1} f(z)dz = F(z_1) - F(z_0) \ \ \frac{dF}{dz}=f(z) \\
\int_{-2\iota} ^{2\iota}z^4dz = \frac{z^5}{5} = \frac{2^6 \iota}{5} \\
Theorem 2: \int_C f(z)dz = \int_a ^bf(z(t)) \frac{dz}{dt}dt\\. The path C can be represented by z(t)=2\exp{\iota t} \ f(z)=(2\exp{\iota t})^4 \ = \ 2^4\exp{4\iota t}\ \mbox{ therefore }\ \frac{dz}{dt}= 2 \iota\exp{\iota t} \\ \ \mbox{ therefore, the integral}\ = 2^5\iota\int_{\frac{-\pi}{2}} ^{\frac{\pi}{2}} \exp{5\iota t}dt \ = \ 2^5\iota\frac{\exp{6it}}{6\iota} \ = \ \frac{2^5}{3}(\frac{\exp{3\pi \iota} - \exp{-3\pi \iota} }{2}) =0 \\. Both theorems should give the same answer, but I don't get the same answer. Where have I gone wrong. Help please. Thanks.
Theorem 1 : \int_{z_0} ^{z_1} f(z)dz = F(z_1) - F(z_0) \ \ \frac{dF}{dz}=f(z) \\
\int_{-2\iota} ^{2\iota}z^4dz = \frac{z^5}{5} = \frac{2^6 \iota}{5} \\
Theorem 2: \int_C f(z)dz = \int_a ^bf(z(t)) \frac{dz}{dt}dt\\. The path C can be represented by z(t)=2\exp{\iota t} \ f(z)=(2\exp{\iota t})^4 \ = \ 2^4\exp{4\iota t}\ \mbox{ therefore }\ \frac{dz}{dt}= 2 \iota\exp{\iota t} \\ \ \mbox{ therefore, the integral}\ = 2^5\iota\int_{\frac{-\pi}{2}} ^{\frac{\pi}{2}} \exp{5\iota t}dt \ = \ 2^5\iota\frac{\exp{6it}}{6\iota} \ = \ \frac{2^5}{3}(\frac{\exp{3\pi \iota} - \exp{-3\pi \iota} }{2}) =0 \\. Both theorems should give the same answer, but I don't get the same answer. Where have I gone wrong. Help please. Thanks.
