ME5332 MIDTERM EXAMINATION

4-5:30 pm, March 22, 2021, Open book/note

1. Obtain the closed form for
   

   sin θ+ sin 2θ+sin 3θ+…+sin (n−1)θ.
2. The imaginary part of an analytic function, f(z), is given by
   

   v = sin(x)  (−ey+e−y).

   1. Find u.
   2. Find f(z) as a function of z alone.
3. Evaluate the following integral
   
   

   ⌠ ⌡ ∞ 0  d x (x+1)(x+2) ,

   by considering
   
   

   ⌠ (⎜) ⌡ log z (z+1)(z+2) d z,

   using the integral path discussed in class. No other method is accepted. You have to show the logic and derivations.
4. 1. Identify the order of singularity at z=0 for
      

      ez − 1 (1 − cos z)3 .
   2. Find z that makes 1^z=−1 possible.
   3. Evaluate
      

      ⌠ (⎜) ⌡ 1 z − 1 dz

      where the integral path is a rectangle whose corners are −1, +2, 2 + i and −1 + i ccw ?
   4. Map a circle defined by |z| = 2 in the z plane to the w plane by w = [1/(z−2)].

Solution

1. Start with the sum of a geometric series up to and including the zⁿ⁻¹ term as
   

   1+ z + z2 +z3 + …+ zn−1 = 1−zn 1−z .

   Note that
   

   sin θ+ sin 2θ+sin 3θ+…+sin (n−1)θ,

   is the imaginary part of
   

   1+eiθ+e2θi+ …+ e(n−1)θi.

   Substitute z=e^iθ into the above to get
   

   1+eiθ+e2θi+ …+ e(n−1)θi = 1−en θi 1−eiθ = (1−en i θ)(1−e−iθ) (1−eiθ)(1−e−iθ) = 1−e−iθ−en i θ+e(n−1)i θ 1−2 cos θ+ 1 = (Real part) + i × sin θ− sin n θ+ sin (n−1)θ 2(1−cos θ) .

   Therefore,
   

   sin θ+ sin 2θ+ …+sin (n−1)θ = sin θ− sin n θ+ sin (n−1)θ 2(1−cos θ) .
2. 1. From u_x=v_y,
      

      ∂u ∂x = −sin x (ey+e−y) ⇒ u=cos(x) (ey+e−y)+ g(y),

      From u_y=−v_x,
      

      cos(x) (ey−e−y) +g′(y) = − (−cos(x) (−e−y+ey)) ⇒ g′(x)=const.

      Therefore,
      

      u = cos(x) (ey+e−y).
   2. 
      

      u + i v = cos(x)(ey+e−y) + i sin(x)(−ey+e−y) = ey (cos(x)− i sin(x))+e−y(cos(x)+i sin(x)) = ey e−ix+e−yeix = e−ix+y+eix −y = e−i(x+iy)+ ei (x+iy) = e−iz+eiz = 2 cos(z).
3. 
   

   ⌠ (⎜) ⌡ log z (z+1)(z+2) dz = 2 πi (Res(eπi)+Res(2eπi)) = 2 πi ⎛ ⎝ πi 1 + log 2 + πi −1 ⎞ ⎠ = −2 πi log 2.

   On the other hand,
   

   ⌠ (⎜) ⌡ log z (z+1)(z+2) dz = ⌠ ⌡ R ϵ  log x (x+1)(x+2) dx + ⌠ ⌡ R  + ⌠ ⌡ ϵ R  log(e2 πix) (x+1)(x+2) dx + ⌠ ⌡ ϵ  = ⌠ ⌡ ∞ 0  log x (x+1)(x+2) dx+ ⌠ ⌡ 0 ∞  log(e2 πix) (x+1)(x+2) dx = ⌠ ⌡ ∞ 0  log x − log x − 2 πi (x+1)(x+2) dx = −2 πi ⌠ ⌡ ∞ 0  1 (x+1)(x+2) dx.

   Therefore,
   

   ⌠ ⌡ ∞ 0  1 (x+1)(x+2) dx = log 2.
4. 1. At z = 0, e^z −1  ∼ z and 1 − cos z  ∼ [(z²)/2] so (e^z − 1)/(1 − cos z)³  ∼ [8/(z⁵)], i.e. the fifth order.
   2. 1^z=−1 is equivalent to e^{2 n πi z}=e^{(π+2 m π)i} from which
      

      z= 1+2m 2n .
   3. Note that z = 1 lies on the integral path so its residue should be counted as one half.
      

      I = πi.
   4. Solving for z yields z = (1 + 2 w)/w, so |z| = 2 is transformed to
      

      | 1 + 2 w w | = 2,

      which is equivalent to
      

      | w + 1 2 | = | w |,

      i.e.
      

      u = − 1 4 .