Complex analysis – Problem solving strategies.

This post is written in spirit of Terrance Tao’s post on problem solving strategies in real analysis. Our emphasis will be on concrete techniques with plenty of examples (many are taken from past prelim questions in UW).

1. Quantitative estimates using Cauchy’s integral formula

The starting point is Cauchy’s integral formula. If f is analytic in a domain \Omega, then

f(z_0) = \frac{1}{2\pi i} \int_{\gamma} \frac{f(z)}{z - z_0}dz,

where \gamma is a positively oriented contour enclosing z_0. (We need that the winding number  n(\gamma, z_0) is 1.) In many applications, \gamma will be a circle. The basic idea is to use bounds for f on \gamma (the “boundary”) to bound f(z). Besides f(z), we can also bound the higher order derivatives, or equivalently the coefficients of the power series centered at z. Explicitly, if f(z) = \sum_{n = 0}^{\infty} a_n(z - z_0)^n, then

a_n = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(z)}{(z - z_0)^{n+1}}dz.

To estimate the integral, we always use the following basic inequality (which is essential when applying the residue theorem):

\left| \int_{\gamma} g(z) dz \right| \leq \int_{\gamma} |g(z)| |dz|.

Here is the basic example:

Example 1.1. (Liouville’s Theorem) If f is a bounded entire function, then f is constant.

Proof. Let f(z) = \sum_{n = 0}^{\infty} a_nz^n and suppose |f| is bounded by M everywhere. We claim that a_n = 0 for n \geq 1. For any R > 0, applying Cauchy’s integral formula for the circle \partial {\Bbb D}(0, R) gives

a_n = \frac{1}{2 \pi i} \int_{\partial {\Bbb D}(0, R)} \frac{f(z)}{z^{n+1}}dz.


|a_n| \leq \frac{1}{2 \pi} \int_{\partial {\Bbb D}(0, R)} \frac{M}{R^{n+1}} |dz| = \frac{M}{R^n},

which goes to 0 as R \rightarrow \infty. (The last inequality is called Cauchy’s estimate.) \Box

The proof of Liouville’s Theorem can be modified to yield a more quantitative statement.

Example 1.2. Let f be entire. If there exists \alpha > 0 such that |f(z)| \leq C(1 + |z|)^{\alpha}, then f is a polynomial with degree at most \alpha.

Proof. Again we write f(z) = \sum_{n = 0}^{\infty} a_nz^n. By Cauchy’s integral formula,

|a_n| \leq C\frac{(1 + R)^{\alpha}}{R^n}

which goes to 0 as R \rightarrow \infty for n > \alpha. Hence f(z) = \sum_{n = 0}^d a_nz^n, where d is the integer part of \alpha. \Box

In the above examples we let R \rightarrow \infty to obtain the desired result. Sometimes, after apply Cauchy’s integral formula, we need to choose the best R to optimize the estimate.

Example 1.3. (Bloch functions) A holomorphic function f(z) = \sum_{n = 0}^{\infty} a_nz^n on {\Bbb D} is called a Bloch function if

\|f\|_B = \sup_{|z| < 1} (1 - |z|^2) |f'(z)| < \infty.

Prove that there is a positive constant C, independent of f, so that

\sup_{n \geq 1} |a_n| \leq C \|f\|_B.

Proof. Since we have a condition on f', it is natural to express a_n in terms of f'. Since a_n = \frac{f^{(n)}(0)}{n!} = \frac{1}{n!}\frac{(f')^{n-1}(0)}{(n-1)!}, for 0 < r < 1 we have

|a_n| = \left| \frac{1}{2 \pi i n} \int_{\partial {\Bbb D}} \frac{f'(z)}{z^n} dz \right| \leq \frac{1}{2 \pi n} \int_{\partial {\Bbb D}(0, r)} \left| \frac{f'(z)(1 - |z|^2)}{z^n(1 - |z|^2)} \right| |dz|,

which gives

|a_n| \leq \frac{1}{r^{n-1}(1 - r^2)} \|f\|_B.

We need to prove that

\sup_{n \geq 1} \inf_{r \in (0, 1)} \frac{1}{r^{n-1}(1 - r^2)} < \infty.

For each n, the infimum is attained (by calculus) at r = \frac{n-1}{n+1}. Since

\inf_{r \in (0, 1)} \frac{1}{r^{n-1}(1 - r^2)} = \frac{n+1}{2n} \left( \frac{n-1}{n+1}\right)^{-(n-1)} \rightarrow \frac{e^2}{2},

the sequence \inf_{r \in (0, 1)} \frac{1}{r^{n-1}(1 - r^2)} is bounded above. We are done. \Box

For functions defined in a disc {\Bbb D}(0, R), the circle \partial {\Bbb D}(0, r) is the most natural contour. Sometimes it is better to use other contours.

Example 1.4. (Problem 5.4.22 in Berkeley Problems in Mathematics) Let f be analytic on {\Bbb D} such that |f(z)| \leq C/(1 - |z|) for all z \in {\Bbb D}. Prove that |f'(z)| \leq 4C/(1 - |z|)^2.

Proof. Fix z \in {\Bbb D}. Instead of using \partial {\Bbb D}(0, r) for |z| < r < 1, we use the contour \partial {\Bbb D}(z, (1 - |z|)/2). Applying Cauchy’s integral formula, we have

|f'(z)| = \left| \frac{1}{2 \pi i} \int_{\partial {\Bbb D}(z, (1-|z|)/2)} \frac{f(\zeta)}{(\zeta - z)^2} d\zeta \right| \leq \frac{1}{2\pi} \int_{\partial {\Bbb D}(z, (1-|z|)/2)} \frac{C}{1 - |\zeta|} \frac{1}{|\zeta - z|^2} |d\zeta|.

Now on the circle |\zeta| is at most |z| + (1 - |z|)/2. Plugging this in and after some algebra, we get |f'(z)| \leq \frac{4C}{(1 - |z|)^2}. \Box

We give an example were the appropriate contour is not a circle.

Example 1.5. (Problem 5.5.13 of Berkeley Problems in Mathematics) Let the function f be entire and satisfy the inequality (with z = x + iy)

|f(z)| \leq \frac{1}{\sqrt{|x|}}

off the imaginary axis. Prove that f is constant.

Proof. By Liouville’s theorem, it suffices to prove that f is bounded. From the hypothesis, f(z) is bounded by 1 for |x| \geq \sqrt{2}. Fix z with |x| < \frac{1}{2}. Consider the the square contour \partial S with vertices at i\text{Im} z \pm 1 \pm i (the reason will become clear later). By Cauchy’s integral formula,

|f(z)| = \left| \frac{1}{2 \pi i}\int_{\partial S} \frac{f(\zeta)}{\zeta - z} d\zeta \right|.

Since |\zeta - z| \geq \frac{1}{2} for \zeta \in \partial S, it suffices to bound \int_{\partial S} |f(\zeta)| |d\zeta|. Since f is bounded for |x| = 1, the integral on each vertical segment is bounded by 2. For each horizontal segment, the integral is bounded above by

\int_{-1}^1 \frac{1}{|x|^{1/2}} dx < \infty.

Hence f is bounded. \Box

Finally we give a simple but more involved example.

Example 1.6. Let F be holomorphic on a neighborhood of \overline{{\Bbb D}}, and let f be holomorphic on {\Bbb D}. Prove that

\lambda_F(f) = \lim_{r \uparrow 1} \int_0^{2 \pi} \overline{F(re^{i\theta})}f(e^{i\theta})d\theta

exists. Show that, with F fixed, if \{f_n\} is a sequence of functions holomorphic on {\Bbb D} and if f_n \rightarrow g uniformly on compacts – whence g is holomorphic on {\Bbb D} – then \lambda_F(f_n) \rightarrow \lambda_F(g).

Proof. Write F(z) = \sum_{n = 0}^{\infty} a_n z^n and f(z) = \sum_{n = 0}^{\infty} b_n z^n. By assumption and Hadamard’s formula, we have

\limsup_{n \rightarrow \infty} |a_n|^{1/n} < 1, \limsup_{n \rightarrow \infty} |b_n|^{1/n} \leq 1.

For 0 < r < 1, by uniform convergence, we have

\int_0^{2\pi} \overline{F(re^{i\theta})}f(re^{i\theta})d\theta = 2 \pi \sum_{k = 0}^{\infty} \overline{a_k}b_k r^{2k}.

(Here we use the standard fact that \int_0^{2\pi} e^{ik \theta} d\theta = 2 \pi \delta_{0k}.) Consider the last power series. Since

\limsup_{k \rightarrow \infty} |\overline{a_k}b_k|^{1/k} \leq \limsup_{k \rightarrow \infty} |a_k|^{1/2k} \limsup_{k \rightarrow \infty} |b_k|^{1/k} < 1,

the power series convergence for |r| \leq 1. Hence the limit as r \uparrow 1 exists, and equals

\sum_{k = 0}^{\infty} \overline{a_k}b_k.

We note that the series converges absolutely.

Now suppose that f_n \rightarrow g locally uniformly. Write f_n(z) = \sum_{k = 0}^{\infty} b_k^{(n)} z^k, and we want to show that

\lim_{n \rightarrow \infty} \sum_{k = 0}^{\infty} \overline{a_k} b_k^{(n)} = \sum_{k = 0}^{\infty} \overline{a_k} b_k.

Choose r, s > 0 with \limsup_{n \rightarrow \infty} |a_n|^{1/n} < r < s < 1. Using Cauchy’s integral formula and integrating along the circle |z| = s, we have the estimate

|b^{(n)}_k - b_k| \leq \frac{\|f_n - g\|_s}{s^k},

where \|f_n - g\|_s = \max_{|z| = s} |f_n(z) - g(z)|. Next consider estimating the difference

\left| \sum_{k = 0}^{\infty} \overline{a_k}b_k^{(n)} - \sum_{k = 0}^{\infty} \overline{a_k}b_k \right| = \sum_{n = 0}^{\infty} |a_k||b_k - b_k|.

Let \varepsilon > 0 be given. Since \limsup_{n \rightarrow \infty} |a_n|^{1/n} < r and f_n \rightarrow g uniformly on |z| = r, we may choose an integer N so that |a_k| \leq r^k for k \geq N and \|f_n - g\|_r \leq \varepsilon. Now we split the sum into two parts:

\sum_{k = 0}^{N-1} |a_k||b_k^{(n)} - b_k| + \sum_{k = N}^{\infty} |a_k||b_k^{(n)} - b_k|.

Now since b_k^{(n)} \rightarrow b_k, the first sum (having finitely many terms) tends to 0 as n \rightarrow \infty. The second sum is bounded above by

\sum_{k = N}^{\infty} r^k \frac{\varepsilon}{s^k} \leq \frac{s}{s - r}\varepsilon.

Letting n \rightarrow \infty, we have

\limsup_{n \rightarrow \infty} |\lambda_F(f_n) - \lambda_F(g)| \leq \frac{s}{s - r}\varepsilon.

Since \varepsilon > 0 is arbitrary, we are done. \Box

Quantitative estimates can be derived in other ways, say using Schwarz’s lemma (related to conformal mappings).

2. Functions in the disc

The unit disc {\Bbb D} is one of the most fundamental domains. For example, the Riemann mapping theorem states that any simply connected domain not equal to the whole plane is conformally equivalent to {\Bbb D}.  Many things for the unit disc are known explicitly (e.g. the Poisson kernel) . Moreover, every automorphism of {\Bbb D} is a Mobius map:

\varphi(z) = e^{i\theta} \frac{z - a}{1 - \overline{a}z}.

Using a Mobius map, we can move any point to the origin, where the analysis is usually simpler. We illustrate this with the proof of Pick’s lemma, which is the “invariant form” of Schwarz’s lemma.

Example 2.1. (Pick’s lemma) Let f: {\Bbb D} \rightarrow {\Bbb D} be analytic. Then, for z, w \in {\Bbb D} (with z \neq w), we have

\left| \frac{f(z) - f(w)}{1 - \overline{f(w)}f(z)} \right| \leq \left| \frac{z - w}{1 - \overline{w}z}\right| .

Moreover, equality holds for some (and hence every) pair if and only if f is a Mobius map.

Proof. Fix z \neq w in {\Bbb D}. Let \varphi and \psi be Mobius maps defined by \varphi(\zeta) = \frac{\zeta + w}{1 + \overline{w}\zeta} and \psi(\zeta) = \frac{\zeta - f(w)}{1 - \overline{f(w)}\zeta}.  Then \psi\circ f \circ \varphi: {\Bbb D} \rightarrow {\Bbb D} is analytic and sends 0 to 0. By Schwarz’s lemma, for any \zeta \in {\Bbb D} we have

| \psi \circ f \circ \varphi (\zeta) | \leq |\zeta|.

Now letting \zeta = \varphi^{-1}(z) gives the result. As in Schwarz’s lemma, equality holds if and only if \psi \circ f \circ \varphi is a Mobius map.  \Box

Pick’s lemma can be interpreted in terms of the pseudohyperbolic distance \rho(z, w) = \left| \frac{z - w}{1 - \overline{w}z} \right| on {\Bbb D}. In this terminology, Pick’s lemma says that any analytic f: {\Bbb D} \rightarrow {\Bbb D} is Lipschitz in \rho, and it is an isometry if and only if f is a Mobius map. From this viewpoint the following result is “obvious”.

Example 2.2. Let f: {\Bbb D} \rightarrow {\Bbb D} be analytic. If f has two fixed points, then f is the identity.

Proof. Since \rho(f(z_1), f(z_2)) = \rho(z_1, z_2), f is a Mobius map, and a simple calculation shows that f is the identity. \Box

Here is another example where it is useful to move a point to the origin.

Example 2.3. Let \Omega be a bounded domain, and let f: \overline{\Omega} \rightarrow {\Bbb D} be analytic with |f(z)| = 1 for z \in \partial \Omega. Show that if f is non-constant, then f(\Omega) = {\Bbb D}.

Proof. Suppose f is non-constant. By the maximum principle, f(\Omega) \subset {\Bbb D}. Suppose on the contrary that f omits the value w in {\Bbb D}. Let \varphi(z) = \frac{f(z) - w}{1 - \overline{w}f(z)}. Then \varphi \circ f: \Omega \rightarrow {\Bbb D} omits 0.

Now 1/(\varphi \circ f) is analytic on \overline{\Omega} and is unimodular on \partial \Omega. By the maximum principle, |1/(\varphi \circ f)| on \Omega. Now |\varphi \circ f| = 1 on \Omega by the hypothesis. It follows that \varphi \circ f, and hence f, is constant. This contradiction proves that f(\Omega) = {\Bbb D}.  \Box

Remark 2.4. If f does not vanish on a domain \Omega, there are a few things to try. First, 1/f is analytic (as in the above proof). Second, if \Omega is simply connected we may write f = e^g for some analytic g.

For bounded analytic functions on a unit disc, a very useful technique is the Blaschke product.

Theorem 2.5. Let f be analytic on {\Bbb D}, continuous on {\Bbb D}, such that |f(z)| = 1 for z \in \partial {\Bbb D}. Then f has the form

f(z) = a \prod_{i = 1}^n \frac{z - z_i}{1 - \overline{z_i}z},

where z_1, ..., z_n are the zeros of f on {\Bbb D} (duplicated according to multiplicities), and a is a unimodular constant.

The product above is called a Blaskche product.

Proof. Let B(z) be the product on the right (without the constant). It is easy to verify that B is analytic on {\Bbb D}, continuous on \overline{\Bbb D}, and |B| = 1 on \partial {\Bbb D}. Then f/B is analytic on {\Bbb D}, continuous on \overline{\Bbb D}, and |f/B| = 1 on \partial {\Bbb D}. By the maximum principle, |f/B| \leq 1 on {\Bbb D}. But B/f satisfies the same conditions, and so |B/f| \leq 1 on {\Bbb D}. Hence |f/B| = 1; so f = aB for some |a| = 1. \Box

The proof of the following result is similar.

Theorem 2.6. Let f be analytic on {\Bbb D}, |f| \leq 1, and z_1, ..., z_n are zeros of f (maybe repeated not more than multiplicities). Then

f(z) = \prod_{i = 1}^n \frac{z - z_i}{1 - \overline{z_i}z} g(z),

where g is analytic in {\Bbb D} and |g| \leq 1 on {\Bbb D}.

Example 2.8. Let f be analytic in the closed unit disc, with f(-\log 2) = 0 and |f(z)| \leq |e^z| for all z with |z| = 1. Find the best possible upper bound of |f(\log 2)|.

Proof. The function f(z)/e^z is analytic and bounded by one (maximum principle). By Theorem 2.6, write

\frac{f(z)}{e^z} = \frac{z + \log 2}{1 + (\log 2) z} g(z),

where g is analytic and bounded by 1. It follows that

\left| \frac{f(\log 2)}{e^{\log 2}} \right| = \frac{2 \log 2}{1 + (\log 2)^2} |g(\log 2)| \leq \frac{2 \log 2}{1 + (\log 2)^2}

and so |f(\log 2)| \leq \frac{4 \log 2}{1 + (\log 2)^2}. Equality is attained, for example, when g(z) \equiv 1. \Box

Example 2.7. (Blaschke condition) Let f be analytic and bounded on {\Bbb D}, with zeros a_1, a_2, .... Prove that

\sum_{n = 1}^{\infty} (1 - |a_n|) < \infty.

Proof. Without loss of generality we may assume f is bounded by 1. First, we consider the case where f(0) \neq 0. For each N, by Theorem 2.6 there is an analytic function g with |g| \leq 1 such that

f(z) = \prod_{n = 1}^N \frac{z - a_n}{1 - \overline{a_n}z} g_N(z).

Putting f(0) = 0, we have

|f(0)| = |g_N(0)| \prod_{n = 1}^N |a_n| \leq \prod_{n = 1}^N |a_n|.

Taking logarithm, we have

\sum_{n = 1}^N \log \frac{1}{|a_n|} \leq \log \frac{1}{|f(0)|}.

For 0 < x < 1, we have the inequality \log \frac{1}{x} \geq 1 - x. It follows that

\sum_{n = 1}^N (1 - |a_n|) \leq \frac{1}{|f(0)|}.

Letting N \rightarrow \infty, we have \sum_{n = 1}^{\infty} (1 - |a_n|) < \infty.

Now if f(0) = 0, then we may write f(z) = z^k g(z), where k is the order of the zero at 0, and g is analytic and g(0) \neq 0. Now we may apply the proved result to g. \Box

3. Local behaviors – zeros, poles, and singularities

First we try to understand zeros. If f is holomorphic, non-constant, the zeros are isolated. A rewording is the very important identity principle: if f vanishes on a set with an accumulation point in the domain, then f is identically zero.

Example 3.1. Suppose f is entire and |f(z)| = 1 for all z \in {\Bbb R}. Prove that f(z) = e^{g(z)} where g is entire.

Proof. Recall that if F is holomorhpic on a simply connected domain, then F does not vanish if and only if F = e^G for some holomorphic G. Hence, it suffices to prove that f does not vanish. Consider the function

h(z) = \overline{f(\overline{z})}

which is entire (note that this is the formula for the Schwarz reflection). Then fg is entire and, for $x \in {\Bbb R}$,

f(x)h(x) = f(x)\overline{f(x)} = 1.

Hence f(z)g(z) is identically 1 by the identity principle, and so f never vanishes. \Box

If f(z_0) = 0, then near z_0 we may write

f(z) = (z - z_0)^k g(z)

for some integer k \geq 1 (called the order), where g is analytic near z_0 and g(z_0) \neq 0. For generally, for any z_0, the function f - f(z_0) has a zero at z_0. And to say that f(z) = w is equivalent to saying that f - w has a zero at z. This viewpoint is quite useful. For example, to prove that f is univalent is equivalent to proving that f(z) - f(z_1) has no other zeros other than z_1.

The local behavior of f near z_0 can be decribed by the local mapping theorem. If f(z_0) = w_0 and the order of the zero f - w_0 at z_0 is k, then on a neighborhood of z_0 we can write

f(z) = w + h(z)^k,

where h is conformal and h(0) = 0. (Actually h(z) = (z - z_0)g(z)^{1/k} for an appropriate branch.) A usual way to detect zeros is the argument principle. If f is holomorphic and does not vanish on a contour \gamma, the number of zeros it encloses is (counting multiplicities)

\frac{1}{2\pi i} \int_{\gamma} \frac{f'(z)}{f(z)} dz.

(We only consider simple domains such as discs and rectangles, and topological considerations are not needed.) It can also be interpreted as

\frac{1}{2 \pi i} \int_{f \circ \gamma} \frac{1}{w} dw,

which is the winding number of f \circ \gamma around 0. An application of this idea is the proof of Rouche’s theorem, which states that if |f(z) - g(z)| < |g(z)| on \gamma, then f and g has the same number of zeros inside \gamma.

Example 3.2. Suppose f is analytic on {\Bbb D} and one-to-one on \{z: 1/2 < |z| < 1\}. Prove that f is one-to-one on {\Bbb D}.

Proof. Let w_0 \in {\Bbb C} be fixed. Since f is non-constant, there exists $latex \frac{1}{2} < r < 1$, aribrarily close to 1, such that f(z) \neq w_0 for |z| = r. We want to prove that f - w_0 has at most one zero in {\Bbb D}. By the argument principle, the number of zeros of in |z| < r (written \gamma_r) is

\frac{1}{2 \pi i} \int_{\gamma_r} \frac{f'(z)}{f(z) - f(w_0)} dz = \frac{1}{2 \pi i} \int_{f \circ \gamma_r} \frac{1}{w - w_0} dw.

which is the winding number of f \circ \gamma_r around w_0. Since f is univalent for 1/2 < |z| < 1, the curve f \circ \gamma_r is a Jordan curve. Hence the winding number around any point is either 0 or 1. Letting r \uparrow \infty, we get the desired result. \Box

Example 3.3. Let u be a real-valued non-constant harmonic function on a neighborhood of 0 \in {\Bbb C}. Describe the structure of the level set u^{-1}(u(0)) near 0.

Proof. In a neighorhood of 0, we have u = \text{Re} f where f is some non-constant analytic function. By translation, we may assume f(0) = 0. Let k \geq 1 be the order of the zero (which is unknown based on the given information). By the local mapping theorem, we may write

f(z) = h(z)^k

where h is conformal near 0. Now that u^{-1}(0) is the preimage of the imaginary axis under f. The preimage of i {\Bbb R} under the map z \mapsto z^k is k equally spaced straight lines passing through the origin. Hence u^{-1}(0) is the conformaly image of k equally spaced straight lines passing through 0. \Box

Next we consider the following problem.

Example 3.4. Suppose that f is entire, non-constant and

|f(z)| \leq e^{\sqrt{|z|}} for all z \in {\Bbb C}.

For R > 0, let n(R) equal the number of zeros of f having modulus less than or equal to R. Prove that there exist non-negative constants A and B such that for all R > 0,

n(R) \leq A + B \sqrt{R}.

(An example of such a function is \cos(\sqrt{z}).)

Here the argument principle is not useful because what we have is an upper bound of |f(z)|. Here the appropriate tool is Jensen’s formula.

Jensen’s formula. If f is analytic on |z| = R, and z_1, ..., z_n are the zeros of f in |z| < R (counted according to multiplicities), and f(0) = 0. Then

\log |f(0)| = \sum_{j = 1}^n \log \frac{|z_j|}{R} + \frac{1}{2 \pi } \int_0^{2 \pi} \log |f(Re^{i\theta})| d\theta.

The idea to prove Jensen’s formula is to observe that \log \left| \frac{f(z)}{(z - z_1) \cdots (z - z_n)} \right| is harmonic  and then apply the mean value property. If f(0) = 0, we may write f(z) = z^k g(z), and then we may apply the formula to g.

Proof of Example 3.4. Write f(z) = z^k g(z), where g(0) \neq 0. Fix R > 0 and suppose z_1, ..., z_n \neq 0 are the zeros of f in {\Bbb D}. By Jensen’s formula,

\log |g(0)| = \sum_{j = 1}^n \log \frac{|z_j|}{R} + \frac{1}{2 \pi} \int_0^{2 \pi} \log |g(Re^{i\theta})| d\theta.

Now - \sum_{j = 1}^n \log \frac{|z_j|}{R} = \sum_{j = 1}^n \log \frac{R}{|z_j|} \geq n and

\frac{1}{2 \pi} \int_0^{2 \pi} \log |g(Re^{i\theta})| d\theta \leq \sqrt{R} - k \log R.

Combining, we have, for any R > 0,

n(R) = k + n \leq k - \log |g(0)| - k \log R + \sqrt{R}.

For R \geq 1 we can drop the \log R term, and for R < 1 we simply use a large enough constant term. Explicitly, if we pick A = \max\{n(1), k - \log |g(0)| \} and B = 1, then n(R) \leq A + B\sqrt{R} for all R > 0. \Box

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