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Complex Analysis - Maths KU

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212 Chapter 4 Elementary Functions<br />

– π<br />

2<br />

–2 –1<br />

y<br />

π<br />

2<br />

–<br />

(a) The square S0 π<br />

2<br />

2<br />

1<br />

–1<br />

–2<br />

v<br />

1 2<br />

(b) The image E of S 0<br />

π<br />

2<br />

Figure 4.13 The mapping w = sin z<br />

S 0<br />

E 0<br />

Figure 4.14 The cut elliptical regions E0<br />

and E1<br />

5<br />

0<br />

–5<br />

–2<br />

–2<br />

0<br />

0<br />

2<br />

2<br />

E 1<br />

Figure 4.15 A Riemann surface for<br />

w = sin z<br />

S 1<br />

u<br />

x<br />

(iii) Since the complex sine function is periodic, the mapping w =sinz<br />

isnot one-to-one on the complex plane. Constructing a Riemann<br />

surface, for thisfunction, asdescribed in the Remarksat the end<br />

of Section 2.4 and Section 4.1, will help usvisualize the complex<br />

mapping w =sinz. In order to construct a Riemann surface consider<br />

the mapping on the square S0 defined by −π/2 ≤ x ≤ π/2,<br />

−π/2 ≤ y ≤ π/2. From Example 3, we find that the square S0<br />

shown in color in Figure 4.13(a) maps onto the elliptical region E<br />

shown in gray in Figure 4.13(b). Similarly, the adjacent square S1<br />

defined by π/2 ≤ x ≤ 3π/2, −π/2 ≤ y ≤ π/2, also maps onto E. A<br />

Riemann surface is constructed by starting with two copies of E, E0<br />

and E1, representing the images of S0 and S1, respectively. We then<br />

cut E0 and E1 open along the line segmentsin the real axisfrom<br />

1 to cosh (π/2) and from −1 to− cosh (π/2). Asshown in Figure<br />

4.14, the segment shown in color in the boundary of S0 ismapped<br />

onto the segment shown in black in the boundary of E0, while the<br />

dashed segment shown in color in the boundary of S0 ismapped<br />

onto the dashed segment shown in black in the boundary of E0. In<br />

a similar manner, the segments shown in color in the boundary of<br />

S1 are mapped onto the segments shown in black in the boundary of<br />

E1. Part of the Riemann surface consists of the two elliptical regions<br />

E0 and E1 with the segments shown in black glued together and the<br />

dashed segments glued together. To complete the Riemann surface,<br />

we take for every integer n an elliptical region En representing the<br />

image of the square Sn defined by (2n − 1)π/2 ≤ x ≤ (2n +1)π/2,<br />

−π/2 ≤ y ≤ π/2. Each region En iscut open, asE0 and E1 were,<br />

and En isglued to En+1 along their boundariesin a manner analogousto<br />

that used for E0 and E1. ThisReimann surface, placed in<br />

xyz-space, is illustrated in Figure 4.15.<br />

EXERCISES 4.3 Answers to selected odd-numbered problems begin on page ANS-14.<br />

4.3.1 <strong>Complex</strong> Trigonometric Functions<br />

In Problems 1–8, express the value of the given trigonometric function in the form<br />

a + ib.<br />

1. sin (4i) 2. cos (−3i)<br />

�<br />

π<br />

�<br />

3. cos (2 − 4i) 4. sin + i<br />

4<br />

5. tan (2i) 6. cot (π +2i)<br />

�<br />

π<br />

�<br />

7. sec − i<br />

8. csc (1 + i)<br />

2<br />

In Problems 9–12, find all complex values z satisfying the given equation.<br />

9. sin z = i 10. cos z =4<br />

11. sin z = cos z 12. cos z = i sin z

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