Problem:
π∫−πcosθa+bcosθdθ
Solution:
This exercise use the substitution z=eiθ.
That led to dz=ieiθdθ, and therefore dziz=dθ.
The integration limit becomes a unit circle.
Using the Euler's identity cosθ+isinθ=eiθ=z, we can write trigonometric functions in terms of z
cosθ=eiθ+e−iθ2=z+1z2.
sinθ=eiθ−e−iθ2i=z−1z2i.
Here is one more convenient result for factorizing general quadratic equation:
ax2+bx+c=14a(4ax2+4abx+4ac)=14a((2ax)2+4abx+4ac)=14a((2ax)2+4abx+b2−b2+4ac)=14a((2ax+b)2−b2+4ac)=14a((2ax+b)2−(b2−4ac))=14a((2ax+b)2−(Δ)2)=14a((2ax+b+Δ)(2ax+b−Δ))
Now, we are all set to start solving the problems.
π∫−πcosθa+bcosθdθ=∮z+1z2a+bz+1z2dziz=∮z+1z2a+b(z+1z)dziz=∮z2+12az+b(z2+1)dziz=∮z2+1bz2+2az+bdziz
To move on, we factorize the quadratic expression, computing Δ.
Δ=√(2a)2−4(b)(b)=2√a2−b2
So we move on
π∫−πcosθa+bcosθdθ=∮z2+114b(2bz+2a+2√a2−b2)(2bz+2a−2√a2−b2)dziz=∮z2+11b(bz+a+√a2−b2)(bz+a−√a2−b2)dziz=∮b(z2+1)(bz+a+√a2−b2)(bz+a−√a2−b2)dziz
We have three poles with 1st order: z=0, z=−a+√a2−b2b and z=√a2−b2−ab Only the first and last is within the unit circle.
Next, we compute the residue values using limits
Res[b(z2+1)(bz+a+√a2−b2)(bz+a−√a2−b2)1iz,0]=limz→0(zb(z2+1)(bz+a+√a2−b2)(bz+a−√a2−b2)1iz)=limz→0(b(z2+1)(bz+a+√a2−b2)(bz+a−√a2−b2)1i)=b((0)2+1)(b(0)+a+√a2−b2)(b(0)+a−√a2−b2)1i=b(a+√a2−b2)(a−√a2−b2)1i=b(a2−(a2−b2))1i=bb21i=1b1i
We also have
Res[b(z2+1)(bz+a+√a2−b2)(bz+a−√a2−b2)1iz,√a2−b2−ab]=limz→√a2−b2−ab((z−√a2−b2−ab)b(z2+1)(bz+a+√a2−b2)(bz+a−√a2−b2)1iz)=limz→√a2−b2−ab(1b(bz−(√a2−b2−a))b(z2+1)(bz+a+√a2−b2)(bz+a−√a2−b2)1iz)=limz→√a2−b2−ab(1b(bz+a−√a2−b2)b(z2+1)(bz+a+√a2−b2)(bz+a−√a2−b2)1iz)=limz→√a2−b2−ab(1bb(z2+1)bz+a+√a2−b21iz)=1bb((√a2−b2−ab)2+1)b(√a2−b2−ab)+a+√a2−b21i(√a2−b2−ab)=1bb((√a2−b2−ab)2+1)√a2−b2−a+a+√a2−b21i(√a2−b2−ab)=1bb((√a2−b2−ab)2+1)2√a2−b21i(√a2−b2−ab)=1bb2((√a2−b2−ab)2+1)2√a2−b21i(√a2−b2−a)=1b(√a2−b2−a)2+b22√a2−b21i(√a2−b2−a)=1b(a2−b2)−2a√a2−b2+a2+b22√a2−b21i(√a2−b2−a)=1b2a2−2a√a2−b22√a2−b21i(√a2−b2−a)=1b−2a(√a2−b2−a)2√a2−b21i(√a2−b2−a)=1b−a√a2−b21i
Last, we put these back to the integral, we have
π∫−πcosθa+bcosθdθ=2πi(1b1i+1b−a√a2−b21i)=2πb(1−a√a2−b2)
Q.E.D. This took me 1.5 hours.
π∫−πcosθa+bcosθdθ
Solution:
This exercise use the substitution z=eiθ.
That led to dz=ieiθdθ, and therefore dziz=dθ.
The integration limit becomes a unit circle.
Using the Euler's identity cosθ+isinθ=eiθ=z, we can write trigonometric functions in terms of z
cosθ=eiθ+e−iθ2=z+1z2.
sinθ=eiθ−e−iθ2i=z−1z2i.
Here is one more convenient result for factorizing general quadratic equation:
ax2+bx+c=14a(4ax2+4abx+4ac)=14a((2ax)2+4abx+4ac)=14a((2ax)2+4abx+b2−b2+4ac)=14a((2ax+b)2−b2+4ac)=14a((2ax+b)2−(b2−4ac))=14a((2ax+b)2−(Δ)2)=14a((2ax+b+Δ)(2ax+b−Δ))
Now, we are all set to start solving the problems.
π∫−πcosθa+bcosθdθ=∮z+1z2a+bz+1z2dziz=∮z+1z2a+b(z+1z)dziz=∮z2+12az+b(z2+1)dziz=∮z2+1bz2+2az+bdziz
To move on, we factorize the quadratic expression, computing Δ.
Δ=√(2a)2−4(b)(b)=2√a2−b2
So we move on
π∫−πcosθa+bcosθdθ=∮z2+114b(2bz+2a+2√a2−b2)(2bz+2a−2√a2−b2)dziz=∮z2+11b(bz+a+√a2−b2)(bz+a−√a2−b2)dziz=∮b(z2+1)(bz+a+√a2−b2)(bz+a−√a2−b2)dziz
We have three poles with 1st order: z=0, z=−a+√a2−b2b and z=√a2−b2−ab Only the first and last is within the unit circle.
Next, we compute the residue values using limits
Res[b(z2+1)(bz+a+√a2−b2)(bz+a−√a2−b2)1iz,0]=limz→0(zb(z2+1)(bz+a+√a2−b2)(bz+a−√a2−b2)1iz)=limz→0(b(z2+1)(bz+a+√a2−b2)(bz+a−√a2−b2)1i)=b((0)2+1)(b(0)+a+√a2−b2)(b(0)+a−√a2−b2)1i=b(a+√a2−b2)(a−√a2−b2)1i=b(a2−(a2−b2))1i=bb21i=1b1i
We also have
Res[b(z2+1)(bz+a+√a2−b2)(bz+a−√a2−b2)1iz,√a2−b2−ab]=limz→√a2−b2−ab((z−√a2−b2−ab)b(z2+1)(bz+a+√a2−b2)(bz+a−√a2−b2)1iz)=limz→√a2−b2−ab(1b(bz−(√a2−b2−a))b(z2+1)(bz+a+√a2−b2)(bz+a−√a2−b2)1iz)=limz→√a2−b2−ab(1b(bz+a−√a2−b2)b(z2+1)(bz+a+√a2−b2)(bz+a−√a2−b2)1iz)=limz→√a2−b2−ab(1bb(z2+1)bz+a+√a2−b21iz)=1bb((√a2−b2−ab)2+1)b(√a2−b2−ab)+a+√a2−b21i(√a2−b2−ab)=1bb((√a2−b2−ab)2+1)√a2−b2−a+a+√a2−b21i(√a2−b2−ab)=1bb((√a2−b2−ab)2+1)2√a2−b21i(√a2−b2−ab)=1bb2((√a2−b2−ab)2+1)2√a2−b21i(√a2−b2−a)=1b(√a2−b2−a)2+b22√a2−b21i(√a2−b2−a)=1b(a2−b2)−2a√a2−b2+a2+b22√a2−b21i(√a2−b2−a)=1b2a2−2a√a2−b22√a2−b21i(√a2−b2−a)=1b−2a(√a2−b2−a)2√a2−b21i(√a2−b2−a)=1b−a√a2−b21i
Last, we put these back to the integral, we have
π∫−πcosθa+bcosθdθ=2πi(1b1i+1b−a√a2−b21i)=2πb(1−a√a2−b2)
Q.E.D. This took me 1.5 hours.
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