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math121a-f23:september_20_wednesday [2023/09/21 12:16]
pzhou
math121a-f23:september_20_wednesday [2023/09/21 12:30] (current)
pzhou [Exercises]
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 ===== Method 3: change of variable ===== ===== Method 3: change of variable =====
 Let w=1/zw = 1/z, then we have  Let w=1/zw = 1/z, then we have 
-I = \oint_{|w|=1/10, CW} \frac{1}{(w^{-1}-1)(w^{-1}-2)} d (w^{-1}) = \oint_{|w|=1/10, CW} \frac{-1}{(1-w)(1-2w)} dw = \oint_{|w|=1/10} \frac{1}{(1-w)(1-2w)}} dw  + 
-since the integrand is only singular at w=1,1/2w=1,1/2, and the contour w=1/10|w|=1/10 contains no singularity in its interior, the integral is 0. + 
 + I = \oint_{|w|=1/10, CW} \frac{1}{(w^{-1}-1)(w^{-1}-2)} d (w^{-1}) = \oint_{|w|=1/10, CW} \frac{-1}{(1-w)(1-2w)} dw = \oint_{|w|=1/10} \frac{1}{(1-w)(1-2w)} dw  
 +where in the last step, I changed the orientation of the contour from CWCW to CCWCCW(CCW is by default, hence omited) and add an extra (1)(-1) factor to the integral.  
 + 
 +Since the integrand is only singular at w=1,1/2w=1,1/2, and the contour w=1/10|w|=1/10 contains no singularity in its interior, the integral is 0.  
 + 
 +===== Riemann sphere ===== 
 +It is useful to think of add a point \infty to the complex plane C\C, and think of C{}\C \cup \{\infty\} as a sphere, where \infty is identified with the north pole, 00 with the south pole, the unit circle z=1|z|=1 as the equator.  
 + 
 +The natural coordinate to use near the north pole is w=1/zw=1/z, so that z=z=\infty corresponds to w=0w=0
 + 
 + 
 + 
 +===== Exercises ===== 
 +Let CC be the contour of z=10|z|=10. Consider the following integrals. 
 + 
 +(1) C11+z2dz\oint_C \frac{1}{1+z^2} dz  
 + 
 +(2) (the result for this one is not zero.)  
 +Cz1+z2dz\oint_C \frac{z}{1+z^2} dz  
 + 
 +(3) Cz21+z4dz\oint_C \frac{z^2}{1+z^4} dz  
 + 
 +Apply methods 1,2,3 to the above problems (each method need to be used once) 
 + 
  
  
  
  
math121a-f23/september_20_wednesday.1695323765.txt.gz · Last modified: 2023/09/21 12:16 by pzhou