By Edoardo Sernesi

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**Additional resources for Geometria V.2**

**Sample text**

Then |z|=1 sin z sin z 2π cosh 1 · 2π ≤ dz ≤ max < 2πe. 1 We shall here demonstrate – although it will later follow more systematically – that |z|=1 sin z dz = 2πi. z2 First method. If one already knows a little of calculus of residues, then the task is quite simple: We sin z see that z = 0 is a simple pole of , because sin z has a simple zero for z = 0, and because z2 z = 0 is the only singularity inside the circle |z| = 1. Hence by the theorem of residues, |z|=1 sin z ;0 z2 sin z dz = 2πi res z2 = 2πi lim z→0 sin z = 2πi.

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 Please click the advert Re(sinh(1 + i)) = sinh 1 · cos 1, Im(sinh(1 + i)) = cosh 1 · sin 1. We have ambitions. Also for you. SimCorp is a global leader in ﬁnancial software. At SimCorp, you will be part of a large network of competent and skilled colleagues who all aspire to reach common goals with dedication and team spirit. We invest in our employees to ensure that you can meet your ambitions on a personal as well as on a professional level.

So summing up the solution can be written z= π + 2pπ ± 2 √ π − i ln( 2 + 1) , 2 p ∈ Z. DIVERSE - INNOVATIVE - INTERNATIONAL Please click the advert Are you considering a European business degree? Copenhagen Business School is one of the largest business schools in Northern Europe with more than 15,000 students from Europe, North America, Australia and Asia. Are you curious to know how a modern European business school competes with a diverse, innovative and international study environment? 14 Find all solutions of the equation tan z = 3i, in the form x + iy.