Analítico – sintético

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1

n ifn mod 4∈ {0, 1}

−¹n if n mod 4∈ {2, 3}. Define the partial sums Sk:=Pk

n=0ck. We claim that the following chain of inequalities holds 0≤ S^4k+3< S^a) 4k

b)< S4k+4 c)< S4k+2

d)< S4k+1≤ 1.

To verify this, note that

S4k+3− S^4k= c4k+1+ c4k+2+ c4k+3=− 16k²+ 8k− 1

(4k + 1)(4k + 2)(4k + 3) < 0, hence a) S4k+4− S^4k= c4k+3+ c4k+4=− 1

4k + 3+ 1

4k + 4 < 0, hence c) S4k+2− S^4k+1= c4k+2< 0, hence d).

Relation b) is obvious and so are the upper bound c1= 1 and the non-negativity constraint. We remark that {S^4k+l}^k≥1, l∈ {0, 1, 2, 3} describe bounded and monotone subsequences that converge to some point. Now that|cⁿ| & 0 the difference between S^4k+land S4k+m, l, m ∈ 0, 1, 2, 3 tends to zero, i.e. all subsequences converge to the same limit. Therefore the power series converges also in the case of z = i.

3.2 Analytic functions

Exercise 1. Show that f (z) =|z|²= x²+ y²has a derivative only at the origin.

Solution. The derivative of f at z is given by

f⁰(z) = lim

h→0

f (z + h)− f (z)

h , h∈ C

provided the limit exist. We have f (z + h)− f (z)

h

= |z + h|²− |z|² h

= (z + h)(¯z + ¯h)− z¯z h

= z¯z + h¯z + z¯h + h¯h− z¯z h

= ¯z + ¯h + z¯h h =: D.

If the limit of D exists, it may be found by letting the point h = (x, y) approach the origin (0,0) in the complex plane C in any manner.

1.) Take the path along the real axes, that is y = 0. Then ¯h = h and thus D = ¯z + h + zh

h = ¯z + h + z

and therefore, if the limit of D exists, its value has to be ¯z + z.

2.) Take the path along the imaginary axes, that is x = 0. Then ¯h =−h and thus D = ¯z− h − zh

h = ¯z− h − z and therefore, if the limit of D exists, its value has to be ¯z− z.

Because of the uniqueness of the limit of D, we must have

¯z + z = ¯z− z ⇐⇒ z = −z ⇐⇒ z = 0,

if the limit of D exists. It remains to show that the limit of D exists at z = 0. Since z = 0, we have that D = ¯h and thus the limit of D is 0.

In summary, the function f (z) =|z|²= x²+ y²has a derivative only at the origin with value 0.

Exercise 2. Prove that if bn, an are real and positive and 0 < b = lim bn < ∞, a = lim sup aⁿ then ab = lim sup(anbn). Docs this remain true if the requirement of positivity is dropped?

Solution. Let a = lim sup

n→∞

an<∞. Then there exists a monotonic subsequence {aⁿk} of {aⁿ} that converges to a. Since lim

k→∞b_nk = b, lim

n→∞a_nkb_nk = ab. Hence,{ankb_nk} is a subsequence of anb_nthat converges to ab. So ab≤ lim sup

n→∞

anbn. Hence, lim sup

n→∞

anbn ≥ b lim sup

n→∞

an. Now, let a = lim sup

n→∞

an =∞. Then there exists a subsequence {aⁿk} of {aⁿ} such that lim_k

→∞an_k = a > 0.

And, since lim

n→∞bn > 0, lim

k→∞an_kbn_k = ∞. Hence lim sup

n→∞

anbn = ∞. In this second case, lim sup

n→∞

anbn ≥ b lim sup

n→∞

an.

In both cases, we have established that ab≤ lim sup

n→∞

anbn. Now, since for all n∈ N, aⁿ> 0 and bn> 0, consider lim

n→∞

1 bn

= 1

b. Applying the inequality we have established replacing b_nwith _b¹

n and a_nreplaced with anbn:

lim sup

n→∞

an= lim sup

n→∞

1 bn

(anbn)≥1 blim sup

n→∞

anbn. Rearranging,

lim sup

n→∞

anbn≤ b lim sup

n→∞

an. It follows that ab = lim sup

n→∞

anbnas required.

Now, consider the case where we drop the positivity requirement. Let bn = 0,−¹2, 0,−¹3, 0,−¹4, . . . and note 0 = b = lim

n→∞bn <∞. Also let aⁿ = 0,−2, 0, −3, 0, −4 . . . and note lim sup

n→∞

an = a = 0. In this case, ab = 0 , 1 = lim sup

n→∞

anbn.

Exercise 3. Show that lim n^1/n= 1.

Solution. Let n∈ N. Also let a = n^1/n. Then

a = n^1/n

⇐⇒ log a = log n^1/n

⇐⇒ log a = log n n

⇐⇒ lim_n

→∞log a = lim

n→∞

log n n

Now, let f (x) = log(x)

x . Then lim

n→∞f (x) = 0 by L’Hopital’s Rule. Thus, lim

n→∞

log n n = lim

n→∞f (n) = 0 also. So

nlim→∞a = lim

n→∞n^1/n= 1.

Exercise 4. Show that (cos z)⁰=− sin z and (sin z)⁰= cos z.

Solution. We have by definition

(cos z)⁰= 1 2

e^iz+ e^−iz!0

= i 2

e^iz− e^−iz

=−1 2i

e^iz− e^−iz

=− sin z and similarly

(sin z)⁰= 1 2i

e^iz− e^−iz!₀

= i 2i

e^iz+ e^−iz

= 1 2

e^iz+ e^−iz

= cos z.

Exercise 5. Derive formulas (2.14).

Solution. Not available.

Exercise 6. Describe the following sets: {z : e^z = i}, {z : e^z = −1}, {z : e^z = −i}, {z : cos z = 0}, {z : sin z = 0}.

Solution. Using the definition we obtain

{z : e^z= i} =( 1 2 + 2k

! πi

)

, {z : e^z=−1} = {(1 + 2k) πi} ,

{z : e^z=−i} = (

−1 2+ 2k

! πi

)

, {z : cos z = 0} =( 1 2 + k

! π

) , and

{z : sin z = 0} = {kπ}

where k∈ Z.

Exercise 7. Prove formulas for cos(z + w) and sin(z + w).

Solution. We have

cos(z) = e^iz+ e^−iz 2 sin(z) = e^iz− e^−iz

2i .

1. Claim: cos(z) cos(w)− sin(z) sin(w) = cos(z + w).

Proof:

cos(z) cos(w)− sin(z) sin(w) = e^iz+ e^−iz 2

e^iw+ e^−iw

2 −e^iz− e^−iz 2i

e^iw− e^−iw 2i

= 1

4

e^ize^iw+ e^−ize^−iw +1

4

e^ize^iw+ e^−ize^−iw

= 1

2e^ize^iw+1 2e^−ize^−iw

= 1

2

e^ize^iw+ e^−ize^−iw

= cos(z + w).

2. Claim: sin(z) cos(w) + cos(z) sin(w) = sin(z + w).

Proof:

sin(z) cos(w) + cos(z) sin(w) = e^iz− e^−iz 2i

e^iw+ e^−iw

2 +e^iz+ e^−iz 2

e^iw− e^−iw 2i

= 1

4i

e^ize^iw− e^−ize^−iw + 1

4i

e^ize^iw− e^−ize^−iw

= 1

2ie^ize^iw−1 2e^−ize^−iw

= 1

2i

e^ize^iw− e^−ize^−iw

= sin(z + w).

Exercise 8. Define tan z = _{cos z}^{sin z}; where is this function defined and analytic?

Solution. Since both sin z and cos z are analytic in the entire complex plane, it follows from the discussion in the text following Definition 2.3 that tan z = _{cos z}^{sin z} is analytic wherever cos z , 0. Now, cos z = 0 implies that z is real and equal to an odd multiple of^π₂. Thus let

G≡( (2k + 1)π 2

  k ∈Z

) .

Then tan z is defined and analytic on C−G. If z ∈ G, then cos z = 0 so tan z is undefined on the non-extended complex plane.

Exercise 9. Suppose that zn, z∈ G = C − {z : z ≤ 0} and zⁿ = rne^iθn, z = re^iθwhere−π < θ, θⁿ < π. Prove that if zn→ z then θⁿ→ θ and rⁿ→ r.

Solution. Not available.

Exercise 10. Prove the following generalization of Proposition 2.20. Let G and Ω be open in C and suppose f and h are functions defined on G, g : Ω → C and suppose that f (G) ⊂ Ω. Suppose that g and h are analytic, g⁰(ω) , 0 for any ω, that f is continuous, h is one-one, and that they satisfy h(z) = g( f (z)) for z in G. Show that f is analytic. Give a formula for f⁰(z).

Solution. Not available.

Exercise 11. Suppose that f : G → C is a branch of the logarithm and that n is an integer. Prove that zⁿ = exp(n f (z)) for all z in G.

Solution. Let f (z) be a branch of the logarithm so that e^{f (z)} = z. Let n ∈ Z and consider e^{n f (z)}. In the following cases we shall apply several of the properties of the complex exponential function developed in the discussion on page 38 in the text.

CASE 1: Assume n > 0. Then we note that

e^{n f (z)}= ef (z)+ f (z)+···+ f (z) (n times)

= e^{f (z)}e^{f (z)}· · · e^{f (z)}(n times)

= e^{f (z)}n

= xⁿ.

CASE 2: Assume n < 0. Then let m =−n so that m > 0 and e^{n f (z)}= 1

e^{m f (z)}

= 1 z^m = zⁿ, the middle step following from Case 1.

CASE 3: Assume n = 0. Then e^{n f (z)}= e⁰ = 1 = z⁰= zⁿ.

Exercise 12. Show that the real part of the function z¹² is always positive.

Solution. We know that we can write z = re^iθ ,0,−π < θ < π and Log(z) = ln(r) + iθ. Thus z¹² = eLog^z

1

2 = e¹²Logz= e¹²⁽lnr+iθ)= e¹²lnre^12θi= e¹²lnr cos θ

2



+ i sin θ 2



. Hence,

Re(z) = e¹²lnrcos θ 2



> 0, since e¹²lnr> 0 and cos_θ

2

> 0 since−π < θ < π. Thus, the real part of the function z¹² is always positive.

Exercise 13. Let G = C− {z ∈ R : z ≤ 0} and let n be a positive integer. Find all analytic functions f : G→ C such that z = ( f (z))ⁿfor all z∈ G.

Solution. Let Log(z) be the principal branch, then log(z) = Log(z) + 2kπi for some k ∈ Z. Thus, we can write

f (z) = z^1/n= e^log(z)/n= e⁽Log(z)+2kπi)/n= eLog^(z)/n· e^2kπi/n.

We know that the latter factor are the n-th roots of unity and depend only on k and n. They correspond to the n distinct powers of the expression ζ = e^2πi/n. Therefore, the branches of z^1/non the set U are given by

f (z) = ζ^k· eLog(z)/n,

where k = 0, . . . , n− 1 and therefore they are all constant multiples of each other.

Exercise 14. Suppose f : G→ C is analytic and that G is connected. Show that if f (z) is real for all z in G then f is constant.

Solution. First of all, we can write f : G→ C as

f (z) = u(z) + iv(z)

where u, v are real-valued functions. Since f : G→ C is analytic, that is f is continuously differentiable (Definition 2.3), we have that u and v have continuous partial derivatives. By Theorem 2.29, this implies that u, v satisfy the Cauchy-Riemann equations. That is,

∂u

∂x =∂v

∂y and ∂u

∂y =−∂v

∂x. (3.1)

Since f (z) is real∀z ∈ G, this implies

v(z)≡ 0 and therefore f (z) = u(z). So, since v(z) = 0, we have

∂v

∂x = ∂v

∂y = 0 and by (3.1) we obtain

∂u

∂x =∂u

∂y = 0

and thus u⁰(z) = 0 (see reasoning of equation 2.22 and 2.23 on page 41). Hence f⁰(z) = 0. Since G is connected and f : G→ C is differentiable with f⁰(z) = 0∀z ∈ G, we have that f is constant.

Exercise 15. For r > 0 1et A =n

ω : ω = exp₁

z

 where 0 <|z| < ro

; determine the set A.

Solution. Define the set S ={z : 0 < |z| < r} where r > 0. The image of this set under 1/z is clearly T =

( z : 1

r <|z|

) . To find the image of A is the same as finding the image of T under e^z. Claim: The image of A is C− {0} (thus not depending on r).

To proof the claim, we need to show that for w , 0, the equation e^z = w has a solution z∈ T . Using polar coordinates we can write w =|w|e^iθ. We have to find a complex number z = x + iy such that x²+ y²> ¹_r and e^xe^iy= w =|w|e^iθ. We want e^x =|w| and y = θ + 2kπ, for some k ∈ Z. Using x = log |w| and k  0 such that (log|w|)²+ (θ + 2kπ)²> ¹_r, then we found z = x + iy.

Exercise 16. Find an open connected set G⊂ C and two continuous functions f and g defined on G such that f (z)²= g(z)²= 1− z²for all z in G. Can you make G maximal? Are f and g analytic?

Solution. Not available.

Exercise 17. Give the principal branch of √ 1− z.

Solution. Not available.

Exercise 18. Let f : G→ C and g : G → C be branches of z^aand z^brespectively. Show that f g is a branch of z^a+b and f /g is a branch of z^a−b. Suppose that f (G) ⊂ G and g(G) ⊂ G and prove that both f ◦ g and g◦ f are branches of z^ab.

Solution. Not available.

Exercise 19. Let G be a region and define G^∗ = {z : ¯z ∈ G}. If f : G → C is analytic prove that f^∗: G^∗→ C, defined by f^∗(z) = f ( ¯x), is also analytic.

Solution. Let z = x + iy and let f (z) = u(x, y) + iv(x, y). By assumption f is analytic and therefore u and v have continuous partial derivatives. In addition the Cauchy-Riemann Equations ux = vy and u_y = −vxare satisfied. Since f^∗(z) = f ( ¯x), we have f^∗(z) = u^∗(x, y) + iv^∗(x, y) where u^∗(x, y) = u(x,−y) and v^∗(x, y) =−v(x, −y). Hence, we have u^∗x(x, y) = ux(x,−y) = v^y(x,−y), u^∗y(x, y) =−u^y(x,−y) = v^x(x,−y), v^∗_x(x, y) =−v^x(x,−y) and v^∗y(x, y) = vy(x,−y) and therefore u^∗x= v^∗_yand u^∗_y=−v^∗xso f^∗is analytic.

Exercise 20. Let z₁, z2, . . . , znbe complex numbers such that Re z_k> 0 and Re(zl. . . zk) > 0 for 1≤ k ≤ n.

Show that log(z₁. . . zn) = log z₁+. . . + log zn, where log z is the principal branch of the logarithm. If the restrictions on the z_kare removed, does the formula remain valid?

Solution. Let z1, . . . , zn ∈ C such that Re(z^j) > 0 and Re(z1· · · z^j) > 0 for 1≤ j ≤ n. The proof will be by induction. Consider first the case where n = 2. Let z1, z2 ∈ C as above. Then −^π2 < arg z1 < ^π₂,−^π2 <

arg z2<^π₂ and−^π2 < arg(z1z2) <^π₂. But note arg(z1z2) = arg z1+ arg z2implies that−π < arg z¹+ arg z2< π.

Now, recall

log(z₁z₂) = ln|z¹z₂| + i arg(z¹z₂)

= ln|z¹| + ln |z²| + i arg(z¹) + i arg(z2)

= log(z₁) + log(z₂)

Assume this formula is true for n = k− 1. Let z1, . . . , zk∈ C as above. Then log(z1. . . zk) = log((z1· · · z^k−1)zk)

= log(z1· · · z^k−1) + log zk by the case when n = 2

= log z1+ log z2+. . . + log zk−1+ log zkby the case when n = k− 1

Hence, this is true for all n such that z_i, 1≤ i ≤ n, satisfy the restrictions.

If the restriction on the z_kare removed, does the formula remain valid? Consider the complex numbers z₁ =−1 + 2i, z2=−2 + i. Then z1z₂= 0− 5i. Clearly, z1, z₂, and z₁z₂do not meet the restrictions as stated above. Now,

log(z1) = ln|z¹| + i arg z¹ log(z2) = ln|z²| + i arg z²

Thus log z1+ log z2 = ln√

5 + ln√

5 + i(^3π₂) = ln 5 + i(^3π₂), but note that ^3π₂ < (−π, π) and log(z¹z2) = ln 5 + i(−^π2). Thus log z1+ log z2 ,log(z1z2) where log z is the principal branch of the logarithm. Hence, the formula is invalid.

Exercise 21. Prove that there is no branch of the logarithm defined on G = C− {0}. (Hint: Suppose such a branch exists and compare this with the principal branch.)

Solution. Define the subset ˆG of G by ˆG = C− {z ∈ R : z ≤ 0}. We use the notation Log for the principal part of the log on ˆG, that is

Log(z) = log|z| + i arg(z)

where arg(z)∈ (−π, π). We will prove the statement above by contradiction.

Assume f (z) is a branch of the logarithm defined on G. Then restricting f to ˆG gives us a branch of the log on ˆG. Therefore its only difference to the principal branch is 2πik for some k∈ Z. This yields

f (z) = log|z| + i arg(z) + 2πik

where z∈ ˆG. Since f is analytic in G, it is continuous at−1. But we can check that this is not the case, thus we have derived a contradiction

In document UNIVERSIDAD NACIONAL DE TRUJILLO FACULTAD DE CIENCIAS ECONÓMICAS (página 43-0)