On Bounds of the Sine and Cosine Along Straight Lines on the Complex Plane

Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 19 September 2018 doi:10.20944/preprints201809.0365.v1

Peer-reviewed version available at Acta Universitatis Sapientiae, Mathematica 2019, 11, 371-379; doi:10.2478/ausm-2019-0027

ON BOUNDS OF THE SINE AND COSINE ALONG STRAIGHT LINES ON THE COMPLEX PLANE

FENG QI

Institute of Mathematics, Henan Polytechnic University, Jiaozuo, Henan, 454010, China College of Mathematics, Inner Mongolia University for Nationalities, Tongliao, Inner Mongolia, 028043, China Department of Mathematics, College of Science, Tianjin Polytechnic University, Tianjin, 300387, China E-mail: [email protected], [email protected] URL: https: // qifeng618. wordpress. com

Abstract. In the paper, the author discusses and computes bounds of the sine and cosine along straight lines on the complex plane.

1. Motivations In the theory of complex functions, the sine and cosine on the complex plane C are denoted and deﬁned by eiz − e−iz e−iz + e−iz sin z = and cos z = , 2i 2 where z = x + iy and x, y ∈ R. When z = x ∈ R, these two functions become sin x and cos x which satisfy the periodicity and boundedness sin(x + 2kπ) = sin x, cos(x + 2kπ) = cos x, | sin x| ≤ 1, | cos x| ≤ 1 for k ∈ Z. On the other hand, when z = iy for y ∈ R, e−y − ey e−y + ey sin(iy) = → ±i∞ and cos(iy) = → +∞ 2i 2 as y → ±∞. These imply that the sine and cosine are bounded on the real x-axis, but unbounded on the imaginary y-axis. Motivated by the above boundedness, we naturally guess that the complex functions sin z and cos z for z ∈ C are (1) bounded on all straight lines parallel to the real x-axis, (2) unbounded on all straight lines whose slopes are not horizontal. In this paper, we will verify the above guesses and compute bounds for sin z and cos z on all straight lines parallel to the real x-axis.

2010 Mathematics Subject Classiﬁcation. Primary 33B10; Secondary 30A10. Key words and phrases. bound; sine; cosine; horizontal straight line; vertical straight line; complex plane; slope. This paper was typeset using AMS-LATEX. 1

Peer-reviewed version available at Acta Universitatis Sapientiae, Mathematica 2019, 11, 371-379; doi:10.2478/ausm-2019-0027

2 F. QI

2. Unboundedness of sine and cosine Let y = α + βx for constants α ∈ R and β 6= 0. On this straight line on the complex plane, by the triangle inequality for complex numbers, we have i[x+i(α+βx)] −i[x+i(α+βx)] e − e | sin z| = | sin(x + i(α + βx))| = 2i [ix−(α+βx)] −[ix−(α+βx)] e − e 1 [ix−(α+βx)] −[ix−(α+βx)] = ≥ e − e 2i 2

1 −(α+βx) (α+βx) = e − e → +∞, x → ±∞ 2 and i[x+i(α+βx)] −i[x+i(α+βx)] e + e | cos z| = | cos(x + i(α + βx))| = 2 [ix−(α+βx)] −[ix−(α+βx)] e + e 1 [ix−(α+βx)] −[ix−(α+βx)] = ≥ e − e 2 2

1 −(α+βx) (α+βx) = e − e → +∞, x → ±∞. 2 Consequently, the functions sin z and cos z are not bounded along any straight line whose slope is not horizontal. On the vertical line x = γ for any constant γ ∈ R on the complex plane, by the triangle inequality for complex numbers, we have i(γ+iy) −i(γ+iy) e − e | sin z| = | sin(γ + iy)| = 2i

1 i(γ+iy) −i(γ+iy) 1 −y y ≥ |e | − |e | = |e − e | → +∞, y → ±∞ 2 2 and i(γ+iy) −i(γ+iy) e + e | cos z| = | cos(γ + iy)| = 2

1 i(γ+iy) −i(γ+iy) 1 −y y ≥ |e | − |e | = |e − e | → +∞, y → ±∞. 2 2 Consequently, the functions sin z and cos z are not bounded along any vertical line.

3. Boundedness of the sine On the horizontal straight line y = α for any constant α ∈ R on the complex plane C, by the triangle inequality for complex numbers, we have i(x+iα) −i(x+iα) e − e | sin z| = | sin(x + iα)| = 2i (ix−α) −(ix−α) ix e − e 1 e −ix α = = − e e 2i 2 eα ix 1 e −ix α 1 1 α ≤ + e e = + e 2 eα 2 eα and i(x+iα) −i(x+iα) e − e | sin z| = | sin(x + iα)| = 2i Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 19 September 2018 doi:10.20944/preprints201809.0365.v1

Peer-reviewed version available at Acta Universitatis Sapientiae, Mathematica 2019, 11, 371-379; doi:10.2478/ausm-2019-0027

ON BOUNDS OF SINE AND COSINE ALONG STRAIGHT LINES 3

(ix−α) −(ix−α) ix e − e 1 e −ix α = = − e e 2i 2 eα ix 1 e −ix α 1 1 α ≥ − e e = − e . 2 eα 2 eα Therefore, it follows that 1 1 α 1 1 α − e ≤ | sin(x + iα)| ≤ + e , x, α ∈ R. 2 eα 2 eα When z = 2kπ + iα for k ∈ Z, we have ei(2kπ+iα) − e−i(2kπ+iα) i 1 sin z = sin(2kπ + iα) = = − − eα . 2i 2 eα π When z = 2kπ + 2 + iα for k ∈ Z, we have π ei(2kπ+π/2+iα) − e−i(2kπ+π/2+iα) sin z = sin 2kπ + + iα = 2 2i ei(π/2+iα) − e−i(π/2+iα) e−α + eα 1 1 = = = + eα . 2i 2 2 eα When z = (2k + 1)π + iα for k ∈ Z, we have ei((2k+1)π+iα) − e−i((2k+1)π+iα) sin z = sin((2k + 1)π + iα) = 2i ei(π+iα) − e−i(π+iα) 1 1 i 1 = = eα − = − eα − . 2i 2i eα 2 eα π When z = (2k + 1)π + 2 + iα for k ∈ Z, we have π ei((2k+1)π+π/2+iα) − e−i((2k+1)π+π/2+iα) sin z = sin (2k + 1)π + + iα = 2 2i ei(3π/2+iα) − e−i(3π/2+iα) e−α + eα 1 1 = = − = − + eα . 2i 2 2 eα

4. Boundedness of the cosine On the horizontal straight line y = α for any constant α ∈ R on the complex plane C, by the triangle inequality for complex numbers, we have i(x+iα) −i(x+iα) e + e | cos z| = | cos(x + iα)| = 2 (ix−α) −(ix−α) ix e + e 1 e −ix α = = + e e 2 2 eα ix 1 e −ix α 1 1 α ≤ + e e = + e 2 eα 2 eα and i(x+iα) −i(x+iα) e + e | cos z| = | cos(x + iα)| = 2 (ix−α) −(ix−α) ix e + e 1 e −ix α = = + e e 2 2 eα Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 19 September 2018 doi:10.20944/preprints201809.0365.v1

Peer-reviewed version available at Acta Universitatis Sapientiae, Mathematica 2019, 11, 371-379; doi:10.2478/ausm-2019-0027

4 F. QI

ix 1 e −ix α 1 1 α ≥ − e e = − e . 2 eα 2 eα Therefore, it follows that 1 1 α 1 1 α − e ≤ | cos(x + iα)| ≤ + e , x, α ∈ R. 2 eα 2 eα When z = 2kπ + iα for k ∈ Z, we have ei(2kπ+iα) + e−i(2kπ+iα) 1 1 cos z = cos(2kπ + iα) = = + eα . 2 2 eα π When z = 2kπ + 2 + iα for k ∈ Z, we have π ei(2kπ+π/2+iα) + e−i(2kπ+π/2+iα) cos z = cos 2kπ + + iα = 2 2 ei(π/2+iα) + e−i(π/2+iα) ie−α − ieα i 1 = = = − eα . 2 2 2 eα When z = (2k + 1)π + iα for k ∈ Z, we have ei((2k+1)π+iα) + e−i((2k+1)π+iα) cos z = cos((2k + 1)π + iα) = 2 ei(π+iα) + e−i(π+iα) 1 1 = = − + eα . 2 2 eα π When z = (2k + 1)π + 2 + iα for k ∈ Z, we have π ei((2k+1)π+π/2+iα) + e−i((2k+1)π+π/2+iα) cos z = cos (2k + 1)π + + iα = 2 2 ei(3π/2+iα) + e−i(3π/2+iα) −ie−α + ieα i 1 = = = − − eα . 2 2 2 eα

5. Conclusions On the sloped straight line y = α + βx for α ∈ R and β 6= 0 on the complex plane C, the trigonometric functions sin z = sin(x + i(α + βx)) and cos z = cos(x + i(α + βx)) are unbounded. On the vertical line x = γ for any scalar γ ∈ R on the complex plane C, the trigonometric functions sin z = sin(γ + iy) and cos z = cos(γ + iy) are unbounded. On the horizontal line y = α for any constant α ∈ R on the complex plane C, the trigonometric functions sin z = sin(x + iα) and cos z = cos(x + iα) are bounded by the double inequalities | sinh α| ≤ | sin(x + iα)| ≤ cosh α, x, α ∈ R (1) and | sinh α| ≤ | cos(x + iα)| ≤ cosh α, x, α ∈ R (2) whose equalities are attained at points π 3π 2kπ + iα, 2kπ + + iα, (2k + 1)π + iα, 2kπ + + iα 2 2 with concrete values 3π sin(2kπ + iα) = cos 2kπ + + iα = i sinh α, 2 Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 19 September 2018 doi:10.20944/preprints201809.0365.v1

Peer-reviewed version available at Acta Universitatis Sapientiae, Mathematica 2019, 11, 371-379; doi:10.2478/ausm-2019-0027

ON BOUNDS OF SINE AND COSINE ALONG STRAIGHT LINES 5

π sin 2kπ + + iα = cos(2kπ + iα) = cosh α, 2 π sin((2k + 1)π + iα) = cos 2kπ + + iα = −i sinh α, 2 3π sin 2kπ + + iα = cos((2k + 1)π + iα) = − cosh α 2 for k ∈ Z. Letting α → 0 in the double inequalities (1) and (2) leads to 0 ≤ | sin x| ≤ 1 and 0 ≤ | cos x| ≤ 1, x ∈ R. On the horizontal belt zones 0 ≤ A ≤ y ≤ B and −B ≤ y ≤ −A ≤ 0 on the complex plane C, the trigonometric functions sin z = sin(x + iy) and cos z = cos(x + iy) are bounded by the double inequality | sinh A| ≤ | cos(x ± iy)| ≤ cosh B, x ∈ R.