(1) f is analytic at z = 0 and f0(0

SOLUTIONS FOR MIDTERM OF COMPLEX ANALYSIS (2004) 1.

(1) Since ∇²u = ^∂_∂x^2u2 +^∂_∂y^2u2 = −6y + 6y = 0 ⇒ u is harmonic.

(2) ^∂v_∂x = −^∂u_∂y = −3y²+ 3x², ^∂v_∂y = ^∂u_∂x = −6xy ⇒ v(x, y) = x³− 3xy²+ C.

2.

(1) f is analytic at z = 0 and f⁰(0) = 2 6= 0 ⇒ f is conformal at z = 0.

(2)

cos(z + 2π) = e^i(z+2π)+ e^−i(z+2π)

2 = e^2πie^iz+ e^−2πie^−iz

2 = e^iz+ e^−iz

2 = cos z.

3.

(1) (√

3 + i)³(1 −√ 3i)²

= [2(^√₂³ +¹₂i)[2(¹₂ −^√₂³i)]²

= 2⁵[cos^π₆ + i sin^π₆]³[cos^π₃ − i sin^π₃]²

= 2⁵[cos^π₂ + i sin^π₂][cos^2π₃ − i sin^2π₃ ]

= 2⁵i(−¹₂ −^√₂³i)

= 16√

3 − 16i.

(2) (−1)ⁱ= e^{i log(−1)} = ei(i(2n+1)π) = e^−(2n+1)π. 4.

(1) Since_z2^e−9^z is analytic at |z| ≤ 2 ⇒H _ez

(z+3)(z−3)dz = 0.

(2) H _ez

(z+3)(z−3)dz =H ^ez

z+3

z−3dz = 2πi₃₊₃^e3 = ^πi₃e³. 5.

(1) γ(t) =

½ e^i(π2−tπ) 0 ≤ t ≤ ¹₂ 2t + (2t − 1)i ¹₂ ≤ t ≤ 1 (2) f (x + iy) = x²+ iy²,

z(t) = (1 − t) + ti, z⁰(t) = −1 + i,

f (z(t)) = (1 − t)²+ it²

⇒R

f (z)dz =R

[(1 − t)²+ it²](−1 + i) dt = −²₃. 6.

If f (z) = u(x, y) + iv(x, y)

g(z) = u²(x, −y) − v²(x, −y) − 2iu(x, −y)v(x, −y)

∂(Re g)

∂x = _∂x^∂u²(x, −y) −_∂x^∂ v²(x, −y)

= 2u(x, −y)_∂x^∂ u(x, −y) − 2v(x, −y)_∂x^∂v(x, −y)

= 2u(x, −y)_∂x^∂ v(x, −y) + 2v(x, −y)_∂x^∂ u(x, −y)

= _∂y^∂[−2u(x, −y)v(x, −y)]

= ^{∂(Im g)}_y .

Similarly ^{∂(Re g)}_∂y = −^{∂(Im g)}_∂x ⇒ g is analytic.

7.

|e^iz| = |e^ix−y| = |e^−ye^ix| = |e^−y| = e^−y ≤ e². 8.

Since f = u + iv is analytic

⇒ ^∂u_∂x = ^∂v_∂y and ^∂u_∂y = −_∂x^∂v

1

2

⇒ ^∂f_∂z = ¹₂[^∂f_∂x+ i^∂f_∂y]

= ¹₂[^∂u_∂x+ i^∂v_∂x+ i(^∂u_∂y + i^∂v_∂y)]

= ¹₂[^∂u_∂x−^∂v_∂y] +¹₂i[_∂x^∂v+ ^∂u_∂y]

= 0.

9.

Since ^{f (z)}_zn is an entire function and |^{f (z)}_zn | ≤ M

by Liouville’s theorem ⇒ ^{f (z)}_zn is a constant, i.e. ^{f (z)}_zn = c

⇒ f (z) = czⁿ. 10.

f is analytic in D

then the function ¹_f is analytic in D and |_f¹| ≥ 0

Apply the maximum modulus theorem to the function _f¹ ,

|_{f (z)}¹ | attains its maximum value on the boundary.

i.e., |f (z)| attains its minimum value on the boundary.