Section 25

The Geometric Meaning of the Derivative

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Problem 25.1

Let a ray {z:arg⁡(z−z0)=φ}\left\{ z: \arg \left(z-z_{0}\right) = \varphi \right\} emanating from the point z0z_{0} be given. Find the coefficient of linear stretching R(φ)R(\varphi ) at the point z0z_{0} and the angle of rotation α(φ)\alpha (\varphi ) at the point z0z_{0} for this ray under the following mappings:

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(1)

w=z2,z0=1w = z^{2}, z_{0} = 1;

(2)

w=zˉ2,z0=iw = \bar{z}^{2}, z_{0} = i;

(3)

w=ie2z,z0=0w = i e^{2 z}, z_{0} = 0;

(4)

w=2z+izˉ,z0=0w = 2 z+i \bar{z}, \quad z_{0} = 0;

(5)

w=z−z0z+z0(z0≠0)w = \frac{z-z_{0}}{z+z_{0}}\left(z_{0} \neq 0\right);

(6)

w=1−iz1+iz,z0=−iw = \frac{1-i z}{1+i z}, \quad z_{0} = -i.

Problem 25.2

Find the sets of all points z0z_{0} at which the coefficient of linear stretching equals one for the following mappings:

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(1)

w=z2w = z^{2};

(2)

w=z3w = z^{3};

(3)

w=z2−2zw = z^{2}-2 z;

(4)

w=1zw = \frac{1}{z};

(5)

w=1+iz1−izw = \frac{1+i z}{1-i z}.

Problem 25.3

Find the set of all points z0z_{0} at which the angle of rotation is zero for the following mappings:

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(1)

w=iz2w = i z^{2};

(2)

w=−z3w = -z^{3};

(3)

w=z2−2zw = z^{2}-2 z;

(4)

w=izw = \frac{i}{z};

(5)

w=1+iz1−izw = \frac{1+i z}{1-i z}.

Problem 25.4

Let the function w(z)w(z) be regular at the point z0z_{0}, and let the smooth curves Γ1\Gamma_{1} and Γ2\Gamma_{2}, passing through the point z0z_{0}, have the property that

Re(w(z))=Re(w)(z0)(z∈Γ1)Im(w(z))=Im(w)(z0)(z∈Γ2) \begin{array}{ll} \mathrm{Re}\left(w(z)\right) = \mathrm{Re}\left(w\right)\left(z_{0}\right) & \left(z \in \Gamma _{1}\right) \\ \mathrm{Im}\left(w(z)\right) = \mathrm{Im}\left(w\right)\left(z_{0}\right) & \left(z \in \Gamma _{2}\right) \end{array}

Prove that if w′(z0)≠0w^{\prime }\left(z_{0}\right) \neq 0, then the curves Γ1\Gamma_{1} and Γ2\Gamma_{2} intersect at a right angle.

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Problem 25.5

Let the function w(z)w(z) be regular at the point z0z_{0}, and let the smooth curves Γ1\Gamma_{1} and Γ2\Gamma_{2}, passing through the point z0z_{0}, have the property that

∣w(z)∣=∣w(z0)∣(z∈Γ1);Re(w(z))=Re(w)(z0)(z∈Γ2). \begin{aligned} \left|w(z)\right| & = \left|w\left(z_{0}\right)\right| \quad \left(z \in \Gamma _{1}\right) ; \\ \mathrm{Re}\left(w(z)\right) & = \mathrm{Re}\left(w\right)\left(z_{0}\right) \quad \left(z \in \Gamma _{2}\right). \end{aligned}

Prove that if w′(z0)≠0w^{\prime }\left(z_{0}\right) \neq 0, then the curves Γ1\Gamma_{1} and Γ2\Gamma_{2}, upon intersecting at the point z0z_{0}, form angles ±arg⁡w(z0)+kπ\pm \arg w\left(z_{0}\right)+k \pi.

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Problem 25.6

Find the lengths of the images of the following curves Г under the indicated mappings:

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(1)

Γ={z:z=it+1,0⩽t⩽1},w=z2\Gamma = \left\{ z: z = i t+1,0 \leqslant t \leqslant 1\right\} , w = z^{2};

(2)

Γ={z:z=it,0⩽t⩽2π},w=ez\Gamma = \left\{ z: z = i t, 0 \leqslant t \leqslant 2 \pi \right\} , w = e^{z};

(3)

Γ={z:z=(1+i)t,0⩽t⩽2π},w=ez\Gamma = \left\{ z: z = (1+i) t, 0 \leqslant t \leqslant 2 \pi \right\} , w = e^{z};

(4)

Γ={z:z=(1+i)t,0⩽t⩽1},w=zm(m=1,2,…)\Gamma = \left\{ z: z = (1+i) t, 0 \leqslant t \leqslant 1\right\} , w = z^{m}(m = 1,2, \ldots ).

Problem 25.7

Find the areas of the images of the domains DD under the indicated mappings:

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(1)

D={z:2<∣z∣<3,∣arg⁡z∣<π4},w=z2;D = \left\{ z: 2 < \left|z\right| < 3,\left|\arg z\right| < \frac{\pi }{4}\right\} , w = z^{2} ;

(2)

D={z:0<∣Re(z)∣<1,∣Im(z)∣<π},w=ezD = \left\{ z: 0 < \left|\mathrm{Re}\left(z\right)\right| < 1,\left|\mathrm{Im}\left(z\right)\right| < \pi \right\} , w = e^{z};

(3)

D={z:0<∣Re(z)∣<1,1<∣Im(z)∣<2}D = \left\{ z: 0 < \left|\mathrm{Re}\left(z\right)\right| < 1,1 < \left|\mathrm{Im}\left(z\right)\right| < 2\right\}, w=z−1z+1w = \frac{z-1}{z+1}.

Problem 25.8

Find the region onto which the function w=ezw = e^{z} maps the rectangle

D={1<x<2,0<y<8}. D = \left\{ 1 < x < 2,0 < y < 8\right\} .

Compute the area of the region w(D)w(D) using formula (4) and explain why this formula gives an incorrect answer.

Formula (4): if a function ff is regular and univalent in a region DD and maps it onto a region G=f(D)G = f(D), then the area of GG equals S(G)=∬D∣f′(z)∣2 dx dyS(G) = \iint_{D}\left|f^{\prime }(z)\right|^{2} \, dx \, dy.

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Problem 25.9

Let

P(z)=a0+a1z+…+anzn P(z) = a_{0}+a_{1} z+\ldots +a_{n} z^{n}

Denote by L(r)L(r) the length of the image of the circle {z:∣z∣=r}\left\{ z:\left|z\right| = r\right\} under the mapping w=P(z)w = P(z), and by S(r)S(r) the area of the image of the disk {z:∣z∣<r}\left\{ z:\left|z\right| < r\right\} under the same mapping. Prove that:

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(1)

the inequality S(r)⩾πr2∣P′(0)∣2S(r) \geqslant \pi r^{2}\left|P^{\prime }(0)\right|^{2} holds;

(2)

the inequality

∫0rL2(t)tdt⩽2πS(r) \int _{0}^{r} \frac{L^{2}(t)}{t} d t \leqslant 2 \pi S(r)
(3)

the inequality L(r)⩾2πr∣P′(0)∣L(r) \geqslant 2 \pi r\left|P^{\prime }(0)\right| holds.

Hint. First prove the formulas

12π∫02πP′(reiφ)dφ=P′(0);∬∣z∣<r∣P′(z)∣2dxdy=π∑m=0nm∣am∣2r2m \frac{1}{2 \pi } \int _{0}^{2 \pi } P^{\prime }\left(r e^{i \varphi }\right) d \varphi = P^{\prime }(0) ; \quad \iint _{\left|z\right| < r}\left|P^{\prime }(z)\right|^{2} d x d y = \pi \sum _{m = 0}^{n} m\left|a_{m}\right|^{2} r^{2 m}
Problem 25.10

Let Q(z)=z+∑k=1nckz−kQ(z) = z+\sum_{k = 1}^{n} c_{k} z^{-k}. Find the area of the image of the annulus {z:r<∣z∣<R}\left\{ z: r < \left|z\right| < R\right\} under the mapping w=Q(z)w = Q(z), counting the area of each elementary patch centered at the point w0w_{0} as many times as the function Q(z)Q(z) takes the value w0w_{0} in the annulus {z:r<∣z∣<R}\left\{ z: r < \left|z\right| < R\right\}.

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Problem 25.11

Let the function w(z)w(z) be regular in the domain DD, and let its mapping w=w(z)w = w(z) preserve the Euclidean distance between points, i.e., for any pair of points z1∈Dz_{1} \in D and z2∈Dz_{2} \in D the equality holds

∣w(z2)−w(z1)∣=∣z2−z1∣ \left|w\left(z_{2}\right)-w\left(z_{1}\right)\right| = \left|z_{2}-z_{1}\right|

Prove that w(z)=eiφz+aw(z) = e^{i \varphi } z+a, where φ\varphi is a real and aa a complex constant.

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