Complex Functions Examples c-3 - Elementary Analytic by Leif Mejlbro

By Leif Mejlbro

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Com 36 Complex Functions Examples c-3 The exponential function and the logarithm function 2 1 –2 –1 0 1 2 –1 –2 2 Figure 6: Some level curves F (z)| = ex −y 2 = R > 0. (b) By using polar coordinates we get the description F r eiθ = exp r2 cos 2θ · exp i r2 sin 2θ . This represents a negative number, if and only if (1) r2 sin 2θ = π + 2pπ, p ∈ Z. Since r ≥ 0 varies, the only possibility that (1) is never fulfilled is when sin 2θ = 0, thus if and only if θ0 ∈ − π π , 0, , π , 2 2 because we have assumed that θ0 ∈ ] − π, π].

If k = p · π , p ∈ Z, the cos k · sin k = 0, so 2 u = cosh y, cos k v = − sinh y. sin k When we eliminate the parameter y we get an equation of a hyperbola v2 u2 − = 1. cos2 k sin2 k Now let Im(z) = y = k. If k = 0, then x ∈ R, (u, v) = (cos x, 0), and the image is the interval [−1, 1] on the u-axis. If k = 0, then u = cos x, cosh k v = − sin x, sinh k x ∈ R. When we eliminate the parameter x we obtain an equation of an ellipse v2 u2 = 1. 4 Prove that sin z dz ≤ 2πe. z2 |z|=1 It follows from sin z = sin(x + iy) = sin x · cosh y + i cos x · sinh y, that = sin2 x · cosh2 y + cos2 x · sinh2 y = sin2 x · cosh2 y + cos2 x · (cosh2 y − 1) = cosh2 y − cos2 x.

B) Since ei = cos 1 + i sin 1, we get Re ei = cos 1 and Im ei = sin 1. (c) Since sin(πi) = i sinh π, we get Re(sin(πi)) = 0 og Im(sin(πi)) = sinh π. (d) Since sinh(1 + i) = sinh 1 · cos 1 + i cosh 1 · sin 1, we get Re(sinh(1 + i)) = sinh 1 · cos 1, Im(sinh(1 + i)) = cosh 1 · sin 1. Sharp Minds - Bright Ideas! Please click the advert Employees at FOSS Analytical A/S are living proof of the company value - First - using new inventions to make dedicated solutions for our customers. With sharp minds and cross functional teamwork, we constantly strive to develop new unique products Would you like to join our team?

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