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Elementary Functions
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EXAMPLE 46 Show that cos 2 z + sin 2 z = 1 SOLUTION We start by writing z = x + iy and utilize the fact that cos 2 x + sin 2 x = 1, when x is a real variable Using the result of the last example we have sin( x + iy) = sin x cos iy + cos x sin iy cos( x + iy) = cos x cos iy sin x sin iy Therefore sin 2 z = (sin x cos iy + cos x sin iy)2 = sin 2 x cos 2 iy + cos 2 x sin 2 iy + 2 cos x sin x cos iy sin iy cos 2 z = (cos x cos iy sin x sin iy)2 = cos 2 x cos 2 iy + sin 2 x sin 2 iy 2 cos x sin x cos iy sin iy So it follows that cos 2 z + sin 2 z = sin 2 x cos 2 iy + cos 2 x sin 2 iy + 2 cos x sin x cos iy sin iy + cos 2 x cos 2 iy + sin 2 x sin 2 iy 2 cos x sin x cos iy sin iy = cos 2 x cos 2 iy + sin 2 x cos 2 iy + cos 2 x sin 2 iy + sin 2 x sin 2 iy = cos 2 iy (cos 2 x + sin 2 x ) + sin 2 iy (cos 2 x + sin 2 x ) = cos 2 iy + sin 2 iy Now we expand each terms using Euler s identity: ei ( iy ) + e i ( iy ) cos iy = 2
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2 2 2
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e 2 y + e 2 y + 2 e y + e y = = 2 4 and ei ( iy ) e i ( iy ) sin 2 iy = 2i
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e 2 y + e 2 y 2 e y e y = = 2i 4
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Therefore cos 2 iy + sin 2 iy =
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e2 y + e 2 y + 2 e2 y + e 2 y 2 1 1 = 2 + 2 =1 4 4
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2 2 Hence, cos z + sin z = 1 Following real variables, the trigonometric functions of a complex variable have inverses Let z = cos w Then we de ne the inverse w = cos 1 z, which we call the arc cosine function or cosine inverse There is an inverse trigonometric function for each of the trigonometric functions de ned in Eqs (410) (415) The inverses are written in terms of the complex logarithm (see Example 18 for a derivation) The formulas are
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1 cos 1 z = ln ( z + z 2 1 ) i 1 sin 1 z = ln ( iz + 1 z 2 ) i tan 1 z = 1 1 + iz ln 2i 1 iz
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(416) (417) (418)
1 1 + 1 z2 sec 1 z = ln i z 1 i + z2 1 csc 1 z = ln i z cot 1 z = 1 z +i ln 2i z i
(419)
(420)
(421)
The Hyperbolic Functions
The complex hyperbolic functions are de ned in terms of the complex exponential as follows: cosh z = ez + e z 2 (422)
Elementary Functions
2 1 Cosh z 0 1 2 0 025 05 x 5 1 0 y 5
Figure 410 A plot of cosh z with 0 x 1
ez e z sinh z = 2
(423)
These functions show some interesting features Let s take a closer look at cosh z A plot of cosh z is shown in Fig 410 focusing on the region 0 x 1 Note the oscillations along the y direction These oscillations result from the fact that this function has trigonometric functions with y argument To see this, we write the hyperbolic cosine function in terms of z = x + iy: cosh z = ez + e z 2 x + iy e + e ( x+iy ) = 2 x e (cos y + i sin y) + e x (cos y i sin y) = 2 = cos y cosh x + i sin y sinh x
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