Elementary Functions of a Complex Variable

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Find the value of tg⁡πi2\tg \dfrac {\pi i}{2}.

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Find the value of sin⁡(x+iy)\sin (x+iy).

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Find the value of cos⁡(x+iy)\cos (x+iy).

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Compute the values of the function eze^{z} at the points:

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(1)

z=2πiz = 2 \pi i;

(2)

z=πiz = \pi i;

(3)

z=πi/2z = \pi i / 2;

(4)

z=−πi/2z = -\pi i / 2;

(5)

z=πi/4z = \pi i / 4.

Prove the formulas:

?
(1)

cos⁡(−z)=cos⁡z\cos (-z) = \cos z;

(2)

sin⁡(−z)=−sin⁡z\sin (-z) = -\sin z;

(3)

sin⁡(z+π2)=cos⁡z\sin \left(z+\frac{\pi }{2}\right) = \cos z;

(4)

sin⁡(π+z)=−sin⁡z\sin (\pi +z) = -\sin z;

(5)

sin⁡(z+2π)=sin⁡z\sin (z+2 \pi ) = \sin z;

(6)

cos⁡(z+π2)=−sin⁡z\cos \left(z+\frac{\pi }{2}\right) = -\sin z;

(7)

sin⁡(z1+z2)=sin⁡z1cos⁡z2+cos⁡z1sin⁡z2\sin \left(z_{1}+z_{2}\right) = \sin z_{1} \cos z_{2}+\cos z_{1} \sin z_{2};

(8)

cos⁡(z1+z2)=cos⁡z1cos⁡z2−sin⁡z1sin⁡z2\cos \left(z_{1}+z_{2}\right) = \cos z_{1} \cos z_{2}-\sin z_{1} \sin z_{2}.

Prove the formulas:

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(1)

ch⁡(−z)=ch⁡z\operatorname {ch}(-z) = \operatorname {ch} z;

(2)

sh⁡(−z)=−sh⁡z\operatorname {sh}(-z) = -\operatorname {sh} z;

(3)

ch⁡2z−sh⁡2z=1\operatorname {ch}^{2} z-\operatorname {sh}^{2} z = 1;

(4)

ch⁡(z1+z2)=ch⁡z1ch⁡z2+sh⁡z1sh⁡z2\operatorname {ch}\left(z_{1}+z_{2}\right) = \operatorname {ch} z_{1} \operatorname {ch} z_{2}+\operatorname {sh} z_{1} \operatorname {sh} z_{2};

(5)

sh⁡(z+πi2)=ich⁡z\operatorname {sh}\left(z+\frac{\pi i}{2}\right) = i \operatorname {ch} z;

(6)

ch⁡(z+πi2)=ish⁡z\operatorname {ch}\left(z+\frac{\pi i}{2}\right) = i \operatorname {sh} z;

(7)

sh⁡(z+πi)=−sh⁡z\operatorname {sh}(z+\pi i) = -\operatorname {sh} z;

(8)

ch⁡(z+πi)=−ch⁡z\operatorname {ch}(z+\pi i) = -\operatorname {ch} z;

(9)

th⁡(z+πi)=th⁡z\operatorname {th}(z+\pi i) = \operatorname {th} z;

(10)

ch⁡(z+2πi)=ch⁡z\operatorname {ch}(z+2 \pi i) = \operatorname {ch} z.

Prove the formulas:

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(1)

sh⁡z=−isin⁡(iz)\operatorname {sh} z = -i \sin (i z);

(2)

sh⁡(iz)=isin⁡z\operatorname {sh}(i z) = i \sin z;

(3)

cos⁡(iz)=ch⁡z\cos (i z) = \operatorname {ch} z;

(4)

ch⁡(iz)=cos⁡z\operatorname {ch}(i z) = \cos z;

(5)

tg⁡(iz)=ith⁡z\operatorname {tg}(i z) = i \operatorname {th} z;

(6)

th⁡(iz)=itg⁡z\operatorname {th}(i z) = i \operatorname {tg} z;

(7)

ctg⁡(iz)=−icth⁡z\operatorname {ctg}(i z) = -i \operatorname {cth} z;

(8)

cth⁡(iz)=−ictg⁡z\operatorname {cth}(i z) = -i \operatorname {ctg} z.

Let x=Re(z),y=Im(z)x = \mathrm{Re}\left(z\right), y = \mathrm{Im}\left(z\right). Prove that:

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(1)

Re(sin⁡z)=sin⁡x⋅ch⁡y,Im(sin⁡z)=cos⁡x⋅sh⁡y;\mathrm{Re}\left(\sin z\right) = \sin x \cdot \operatorname {ch} y, \quad \mathrm{Im}\left(\sin z\right) = \cos x \cdot \operatorname {sh} y ;

(2)

Re(cos⁡z)=cos⁡x⋅ch⁡y,Im(cos⁡z)=−sin⁡x⋅sh⁡y\mathrm{Re}\left(\cos z\right) = \cos x \cdot \operatorname {ch} y, \quad \mathrm{Im}\left(\cos z\right) = -\sin x \cdot \operatorname {sh} y;

(3)

Re(sh⁡z)=sh⁡x⋅cos⁡y,Im(sh⁡z)=ch⁡x⋅sin⁡y\mathrm{Re}\left(\operatorname {sh} z\right) = \operatorname {sh} x \cdot \cos y, \quad \mathrm{Im}\left(\operatorname {sh} z\right) = \operatorname {ch} x \cdot \sin y;

(4)

Re(ch⁡z)=ch⁡x⋅cos⁡y,Im(ch⁡z)=sh⁡x⋅sin⁡y\mathrm{Re}\left(\operatorname {ch} z\right) = \operatorname {ch} x \cdot \cos y, \quad \mathrm{Im}\left(\operatorname {ch} z\right) = \operatorname {sh} x \cdot \sin y;

(5)

Re(tg⁡z)=sin⁡2xcos⁡2x+ch⁡2y,Im(tg⁡z)=sh⁡2ycos⁡2x+ch⁡2y\mathrm{Re}\left(\operatorname {tg} z\right) = \frac{\sin 2 x}{\cos 2 x+\operatorname {ch} 2 y}, \quad \mathrm{Im}\left(\operatorname {tg} z\right) = \frac{\operatorname {sh} 2 y}{\cos 2 x+\operatorname {ch} 2 y};

(6)

Re(ctg⁡z)=sin⁡2xch⁡2y−cos⁡2x,Im(ctg⁡z)=−sh⁡2ych⁡2y−cos⁡2x\mathrm{Re}\left(\operatorname {ctg} z\right) = \frac{\sin 2 x}{\operatorname {ch} 2 y-\cos 2 x}, \quad \mathrm{Im}\left(\operatorname {ctg} z\right) = \frac{-\operatorname {sh} 2 y}{\operatorname {ch} 2 y-\cos 2 x}.

Let x=Re(z),y=Im(z)x = \mathrm{Re}\left(z\right), y = \mathrm{Im}\left(z\right). Prove that:

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(1)

∣sin⁡z∣=ch⁡2y−cos⁡2x\left|\sin z\right| = \sqrt{\operatorname {ch}^{2} y-\cos^{2} x};

(2)

∣cos⁡z∣=ch⁡2y−sin⁡2x\left|\cos z\right| = \sqrt{\operatorname {ch}^{2} y-\sin^{2} x};

(3)

∣sh⁡z∣=ch⁡2x−cos⁡2y\left|\operatorname {sh} z\right| = \sqrt{\operatorname {ch}^{2} x-\cos^{2} y};

(4)

∣ch⁡z∣=ch⁡2x−sin⁡2y\left|\operatorname {ch} z\right| = \sqrt{\operatorname {ch}^{2} x-\sin^{2} y};

(5)

∣tg⁡z∣=ch⁡2y−cos⁡2xch⁡2y+cos⁡2x\left|\operatorname {tg} z\right| = \sqrt{\frac{\operatorname {ch} 2 y-\cos 2 x}{\operatorname {ch} 2 y+\cos 2 x}};

(6)

∣cth⁡z∣=ch⁡2x+cos⁡2ych⁡2x−cos⁡2y\left|\operatorname {cth} z\right| = \sqrt{\frac{\operatorname {ch} 2 x+\cos 2 y}{\operatorname {ch} 2 x-\cos 2 y}}.

Describe the points zz at which the following functions take real values:

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(1)

cos⁡z\cos z;

(2)

ch⁡z\operatorname {ch} z;

(3)

sin⁡z\sin z;

(4)

tg⁡z\operatorname {tg} z;

(5)

cth⁡z\operatorname {cth} z.

Describe the points zz at which the following functions take purely imaginary values:

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(1)

sin⁡z\sin z;

(2)

sh⁡z\operatorname {sh} z;

(3)

cos⁡z\cos z;

(4)

ctg⁡z\operatorname {ctg} z;

(5)

th⁡z\operatorname {th} z.

Find all points at which the following functions vanish:

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(1)

sin⁡z\sin z;

(2)

cos⁡z\cos z;

(3)

sh⁡z\operatorname {sh} z;

(4)

ch⁡z\operatorname {ch} z.

Find all solutions of the following equations:

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(1)

sin⁡z=4i3\sin z = \frac{4 i}{3};

(2)

sin⁡z=53\sin z = \frac{5}{3};

(3)

cos⁡z=3i4\cos z = \frac{3 i}{4};

(4)

cos⁡z=3+i4\cos z = \frac{3+i}{4};

(5)

tg⁡z=5i3\operatorname {tg} z = \frac{5 i}{3};

(6)

ctg⁡z=−3i5\operatorname {ctg} z = -\frac{3 i}{5};

(7)

sh⁡z=i2\operatorname {sh} z = \frac{i}{2};

(8)

ch⁡z=12\operatorname {ch} z = \frac{1}{2}.

Prove that for ∣z∣⩽R\left|z\right| \leqslant R the following inequalities hold:

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(1)

∣ch⁡z∣⩽ch⁡R\left|\operatorname {ch} z\right| \leqslant \operatorname {ch} R;

(2)

∣sh⁡z∣⩽sh⁡R\left|\operatorname {sh} z\right| \leqslant \operatorname {sh} R;

(3)

∣cos⁡z∣⩽ch⁡R\left|\cos z\right| \leqslant \operatorname {ch} R;

(4)

∣sin⁡z∣⩽sh⁡R\left|\sin z\right| \leqslant \operatorname {sh} R.

Let x=Re(z),y=Im(z)x = \mathrm{Re}\left(z\right), \quad y = \mathrm{Im}\left(z\right). Prove that:

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(1)

The function eze^{z} tends to infinity as xx tends to +∞+\infty, and this tendency is uniform in yy.

(2)

The function eze^{z} tends to zero as x→−∞x \rightarrow -\infty, and this tendency is uniform in yy.

(3)

The functions sin⁡z\sin z and cos⁡z\cos z tend to infinity as y→±∞y \rightarrow \pm \infty, and this tendency is uniform in xx.

Prove that:

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(1)

The function ez2e^{z^{2}} tends to infinity as z→∞z \rightarrow \infty in any angle of the form

∣arg⁡z−π∣⩽α \left|\arg z-\pi \right| \leqslant \alpha

and in any angle of the form ∣arg⁡z∣⩽α\left|\arg z\right| \leqslant \alpha, provided only that α<π4\alpha < \frac{\pi }{4}.

(2)

The function ez2e^{z^{2}} tends to zero as z→∞z \rightarrow \infty in any angle of the form

∣arg⁡z±π2∣⩽α<π4 \left|\arg z \pm \frac{\pi }{2}\right| \leqslant \alpha < \frac{\pi }{4}

Let P(z)=zn+a1zn−1+…+anP(z) = z^{n}+a_{1} z^{n-1}+\ldots +a_{n}. Prove that the function eP(z)e^{P(z)} tends to infinity as z→∞z \rightarrow \infty in the angles

∣arg⁡z−2kπn∣⩽α<π2n(k=0,1,…,n−1), \left|\arg z-\frac{2 k \pi }{n}\right| \leqslant \alpha < \frac{\pi }{2 n} \quad (k = 0,1, \ldots , n-1),

and tends to zero as z→∞z \rightarrow \infty in the angles

∣arg⁡z−(2k+1)πn∣⩽α<π2n(k=0,1,…,n−1). \left|\arg z-\frac{(2 k+1) \pi }{n}\right| \leqslant \alpha < \frac{\pi }{2 n} \quad (k = 0,1, \ldots , n-1).
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Let x=Re(z),y=Im(z)x = \mathrm{Re}\left(z\right), y = \mathrm{Im}\left(z\right). Prove the inequalities:

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(1)

2e−2y1+e−2y⩽∣tg⁡z−i∣⩽2e−2y1−e−2y(y>0)\frac{2 e^{-2 y}}{1+e^{-2 y}} \leqslant \left|\operatorname {tg} z-i\right| \leqslant \frac{2 e^{-2 y}}{1-e^{-2 y}} \quad (y > 0);

(2)

2e2y1+e2y⩽∣tg⁡z+i∣⩽2e2y1−e−2y(y<0)\frac{2 e^{2 y}}{1+e^{2 y}} \leqslant \left|\operatorname {tg} z+i\right| \leqslant \frac{2 e^{2 y}}{1-e^{-2 y}} \quad (y < 0);

(3)

2e−2y1+e−2y⩽∣ctg⁡z+i∣⩽2e−2y1−e−2y(y>0)\frac{2 e^{-2 y}}{1+e^{-2 y}} \leqslant \left|\operatorname {ctg} z+i\right| \leqslant \frac{2 e^{-2 y}}{1-e^{-2 y}} \quad (y > 0);

(4)

2e2y1+e2y⩽∣ctg⁡z−i∣⩽2e2y1−e2y(y<0)\frac{2 e^{2 y}}{1+e^{2 y}} \leqslant \left|\operatorname {ctg} z-i\right| \leqslant \frac{2 e^{2 y}}{1-e^{2 y}} \quad (y < 0).

Compute the integral ∫Czsin⁡z  dz\int_{C} z \sin z \; d z, where CC is the straight-line segment going from the point z=0z = 0 to the point z=iz = i.

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Define the function eze^{z} for any complex value z=x+iyz = x+i y by the equality ez=ex(cos⁡y+isin⁡y)e^{z} = e^{x}(\cos y+i \sin y). Prove that the function eze^{z} is differentiable for any zz and the equality (ez)′=ez\left(e^{z}\right)^{\prime } = e^{z} holds.

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Define the functions sh⁡z,ch⁡z,sin⁡z,cos⁡z\operatorname {sh} z, \operatorname {ch} z, \sin z, \cos z by the equalities

sh⁡z=12(ez−e−z),ch⁡z=12(ez+e−z),sin⁡z=12i(eiz−e−iz),cos⁡z=12(eiz+e−iz). \begin{aligned} \operatorname {sh} z & = \frac{1}{2}\left(e^{z}-e^{-z}\right), & \operatorname {ch} z & = \frac{1}{2}\left(e^{z}+e^{-z}\right), \\ \sin z & = \frac{1}{2 i}\left(e^{i z}-e^{-i z}\right), & \cos z & = \frac{1}{2}\left(e^{i z}+e^{-i z}\right). \end{aligned}

Prove that

?
(1)

(sh⁡z)′=ch⁡z(\operatorname {sh} z)^{\prime } = \operatorname {ch} z;

(2)

(ch⁡z)′=sh⁡z(\operatorname {ch} z)^{\prime } = \operatorname {sh} z;

(3)

(sin⁡z)′=cos⁡z(\sin z)^{\prime } = \cos z;

(4)

(cos⁡z)′=−sin⁡z(\cos z)^{\prime } = -\sin z.

Determine at which points z∈Cz \in \mathbb {C} the functions are differentiable, and find their derivatives:

?
(1)

esh⁡ze^{\operatorname {sh} z};

(2)

cos⁡(2ez)\cos \left(2 e^{z}\right);

(3)

sin⁡zsh⁡z+icos⁡zch⁡z\sin z \operatorname {sh} z+i \cos z \operatorname {ch} z;

(4)

ezz\frac{e^{z}}{z};

(5)

zez\frac{z}{e^{z}};

(6)

sin⁡z1+z2\frac{\sin z}{1+z^{2}}.

Determine where the functions are differentiable, and find their derivatives:

?
(1)

tg⁡z\operatorname {tg} z;

(2)

ctg⁡z\operatorname {ctg} z;

(3)

ez+2ez−2\frac{e^{z}+2}{e^{z}-2};

(4)

1tg⁡z+ctg⁡z\frac{1}{\operatorname {tg} z+\operatorname {ctg} z};

(5)

(ez+e−z)−3\left(e^{z}+e^{-z}\right)^{-3};

(6)

sin⁡zsin⁡z−cos⁡z\frac{\sin z}{\sin z-\cos z}.

Find the antiderivatives of the functions:

?
(1)

eaze^{a z};

(2)

ch⁡az\operatorname {ch} a z;

(3)

sh⁡az\operatorname {sh} a z;

(4)

cos⁡az\cos a z;

(5)

sin⁡az\sin a z;

(6)

eazcos⁡bze^{a z} \cos b z;

(7)

zeazz e^{a z};

(8)

z2ch⁡azz^{2} \operatorname {ch} a z;

(9)

zcos⁡azz \cos a z.

Using formulas (3) and (5), prove that:

Formula (3): ez=∑n=0∞znn!,∣z∣<+∞e^{z} = \sum_{n = 0}^{\infty } \frac{z^{n}}{n!}, \quad \left|z\right| < +\infty.

Formula (5): cos⁡z=∑n=0∞(−1)nz2n(2n)!,∣z∣<+∞\cos z = \sum_{n = 0}^{\infty } \frac{(-1)^{n} z^{2 n}}{(2 n)!}, \quad \left|z\right| < +\infty.

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(1)

14(ez+e−z+2cos⁡z)=∑n=0∞z4n(4n)!\frac{1}{4}\left(e^{z}+e^{-z}+2 \cos z\right) = \sum_{n = 0}^{\infty } \frac{z^{4 n}}{(4 n)!};

(2)

13(ez+2e−z/2cos⁡z32)=∑n=0∞z3n(3n)!\frac{1}{3}\left(e^{z}+2 e^{-z / 2} \cos \frac{z \sqrt{3}}{2}\right) = \sum_{n = 0}^{\infty } \frac{z^{3 n}}{(3 n)!}.

Expand the functions in a Taylor series in a neighborhood of the point z=0z = 0:

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(1)

sin⁡2z\sin^{2} z;

(2)

cos⁡3z\cos^{3} z;

(3)

sin⁡4z+cos⁡4z\sin^{4} z+\cos^{4} z;

(4)

cos⁡2z+ch⁡2z\cos^{2} z+\operatorname {ch}^{2} z;

(5)

e2sin⁡ze^{2} \sin z;

(6)

ch⁡z⋅cos⁡z\operatorname {ch} z \cdot \cos z.

Prove that the functions ez,cos⁡z,sin⁡z,ch⁡z,sh⁡ze^{z}, \cos z, \sin z, \operatorname {ch} z, \operatorname {sh} z, originally defined only for real values of the variable, can be regularly continued to the entire complex plane by expanding these functions in a Taylor series.

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Find the widest possible domain into which the following functions can be regularly continued from the real axis:

?
(1)

tg⁡z\operatorname {tg} z;

(2)

ctg⁡z\operatorname {ctg} z;

(3)

th⁡z\operatorname {th} z;

(4)

e1/cos⁡ze^{1 / \cos z};

(5)

e−tg⁡ze^{-\operatorname {tg} z};

(6)

sin⁡(th⁡z)\sin (\operatorname {th} z);

(7)

cos⁡(ez2)\cos \left(e^{z^{2}}\right);

(8)

th⁡(ez)\operatorname {th}\left(e^{z}\right);

(9)

ctg⁡(ch⁡z)\operatorname {ctg}(\operatorname {ch} z).

Relying on the validity of the formulas given below for real values of the variables, prove their validity also for arbitrary complex values of these variables:

?
(1)

ez1+z2=ez1⋅ez2e^{z_{1}+z_{2}} = e^{z_{1}} \cdot e^{z_{2}};

(2)

cos⁡2z+sin⁡2z=1\cos^{2} z+\sin^{2} z = 1;

(3)

sin⁡2z=2sin⁡zcos⁡z\sin 2 z = 2 \sin z \cos z;

(4)

ch⁡2z=ch⁡2z+sh⁡2z\operatorname {ch} 2 z = \operatorname {ch}^{2} z+\operatorname {sh}^{2} z;

(5)

sin⁡(z1+z2)=sin⁡z1cos⁡z2+cos⁡z1sin⁡z2\sin \left(z_{1}+z_{2}\right) = \sin z_{1} \cos z_{2}+\cos z_{1} \sin z_{2};

(6)

ch⁡(z1+z2)=ch⁡z1ch⁡z2+sh⁡z1sh⁡z2\operatorname {ch}\left(z_{1}+z_{2}\right) = \operatorname {ch} z_{1} \operatorname {ch} z_{2}+\operatorname {sh} z_{1} \operatorname {sh} z_{2};

(7)

cos⁡z1+cos⁡z2=2cos⁡z1+z22cos⁡z1−z22\cos z_{1}+\cos z_{2} = 2 \cos \frac{z_{1}+z_{2}}{2} \cos \frac{z_{1}-z_{2}}{2};

(8)

sh⁡z1+sh⁡z2=2sh⁡z1+z22ch⁡z1−z22\operatorname {sh} z_{1}+\operatorname {sh} z_{2} = 2 \operatorname {sh} \frac{z_{1}+z_{2}}{2} \operatorname {ch} \frac{z_{1}-z_{2}}{2}.

References

[ГюКу03] Гюнтер Н.М., Кузьмин Р.О. Сборник задач по высшей математике. 13-е изд., Издательство «Лань», 2003.
[ШПК15] Шабунин М.И., Половинкин Е.С., Карлов М.И. Сборник задач по теории функций комплексного переменного. 4-е изд., БИНОМ. Лаборатория знаний, 2015.