COMPLEX ANALYSIS

COMPLEX ANALYSIS 

M. Velrajan

1.

 

 2.

1 / (z4 - 1) = f(z) / (z - i),

f(z) =1 / (z2 - 1)(z + i) is analytic on and inside C, by Cauchy’s integral formula,

the required integral = 2πi f(i) = - π/2.        Ans. 1.

3.

Izm,zn = 2𝜋i, if n = m + 1, 0 in all other cases(on the unit circle conjugate of z is 1/z).

Hence 1, 2 are false.  By Cauchy’s theorem, 3 is true,

hence 4 is false(since p(0) need not be 0). Ans. 3.

4.

By Cauchy’s integral formula, the LHS = 2𝜋i ƛ (taking f(z) = ƛ) and 

the RHS = 2𝜋i(1/(1 - 4))   (taking f(z) = 1/(z - 4)).  Hence  ƛ= - ⅓.     Ans. 1.

5.

By Cauchy’s integral formula, required integral = 2𝝅i(e/(1+1)).  Ans. 2.

6.

If f and its conjugate are analytic in a region D then f is constant on D. 

1 is true.

[If u is harmonic in the region G, then it does not have a strong relative maximum or minimum in G.]  

Since u is harmonic, if u(½) ≥ u(z), for all z in D then u(z) = u(½) for all z in D,

i.e. u is constant and hence f is constant on D. 2 is true.

cos z is analytic and not constant on D, but in the power series of f,

an= 0 for all odd n. 3 is false. 

By 4, f ′(a) = 0, for all a in D with |a| ≥ 1/2. 

Since f ′ is analytic on D, f ′(a) = 0 for all a in D. 4 is true.  Ans. 1, 2, 4.

 

7.

Open Mapping Theorem :     A non-constant holomorphic function on an open set is an open mapping.  

Hence 1 is an open set,  4 is an unbounded open set and hence 2, 3 are closed sets.      Ans. 1.

8.

If f is continuous at zj then it is analytic at zj. Hence 1 is true. 

If f is 1 on Ω - E and 0 on E then f is analytic on Ω - E,

f is bounded but not analytic on Ω. Hence 2 is false. 

If in the Laurent series of

f at zj,

am = 0 for m = - 1, - 2, . . . . then the series is the Taylor series,

hence f is analytic at zj.

But if only a-1 = 0 then zj is a singularity. Hence 3 is true, 4 is false.  Ans. 1, 3.

9.

The function z2 is a non constant analytic function and satisfies 1 and

since 1/n converges to 0 and |0| = 0 < 1.

Hence f(z) = z2.

( If two functions f and g are analytic in a region D and agree on a set with an accumulation point in D then f = g throughout D.)

Hence f exists and is unique and is a polynomial.   Ans. 3.

10.

The function z2 is a non constant analytic function and satisfies 1 and

since 1/n converges to 0 and |0| = 0 < 1.

Hence f(z) = z2

( If two functions f and g are analytic in a region D and agree on a set with an accumulation point in D then f = g throughout D.)  

The function g(z) = z / (2 + z) is non constant analytic on the unit disk D and

g(1/n) = 1/(2n+1), hence, if f satisfy f(1/n) = 1/(2n+1) then g = f on D,

but g(-1/n) = -1/(2n-1). Hence f cannot satisfy 2.

Suppose f(0) = 1 and f satisfy 3 or 4. Then taking lim n→∞, we get contradictions. Hence 3 and 4 are false. Ans. 1.

11.

 

12.

If |f(z)| ≤ 1 then f is a bounded entire function, hence f is a constant function,

hence f ′(z) = 0, for all z, 1 is true. 

Since cos z is onto, if f is onto then f(cos z) is also onto. 

If f(z) = z, for all z then f is analytic on C and onto,

but f(ez) = ez is not onto. 

3 is false.

If p(z) = z4 + z + 2 then f(p(0)) = f(p(-1)) = f(2), hence 4 is false.  Ans. 1, 2.

13.

Clearly 1 is true. 

For x ≤ 0, ex ≤ 1, hence 2 is false.

If f -1{z / |z| ≤ R} is not bounded for some R > 0 then

for each K > 0, there are infinitely many z s.t |z| > K, but |f(z)| ≤ R.

Hence there is an unbounded sequence(zn) s.t. |f(zn)| ≤ R, but since f is a polynomial (f(zn)) should be unbounded.

Hence 3 is true. 

(Since for polynomial f,

lim n→∞ |zn| = ∞ ⟹

lim n→∞ |f(zn)| = ∞.) 

Also, if f is a non-constant entire function, then f maps every unbounded sequence in C to an unbounded sequence if and only if f is a polynomial function.

Hence 4 is false.

(or) (-∞, 0] ⊆ g-1{z in C / |z| ≤ R}, for all R ≥ 1, hence 4 is false. Ans. 1, 3.

 

14.

Any constant function on C is in F, hence 1 is false, 2 is true. 

Let f be in F. If f is a constant function then

f(z) = ke0z. 

Suppose f is non constant. 

Let g(z) = f ′(z) / f(z). Then g is a bounded meromorphic, hence its singularity are removable, hence g extends to a bounded entire function, hence g(z) = 𝛼 is a constant function with |𝛼| ≤ 1. Let h(z) = log(f(z)). Then h′(z) = g(z) = 𝛼, hence

h(z) = 𝛼z + c, where c is a constant. Hence

f(z) = 𝛽e𝛼z, where 𝛽 = ec.

Conversely, let f(z) = 𝛽e𝛼z. Then  f is an entire function and |f ′(z)| = |𝛼| |f(z)|.

Hence f is in F, if |𝛼| ≤ 1.  Hence 4 is true, 3 is false.  Ans. 2, 4.

15.

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COMPLEX ANALYSIS