New PDF release: Complex Functions c-1 - Examples concerning Complex Numbers

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Z We then check what the unit circle |w| = 1 is mapped into by the inverse transformation By putting w = z= 1 . e. w = 1, then z e−iθ − 1 cos θ − 1 − i sin θ cos θ − 1 − i sin θ 1 = iθ = = −iθ iθ −iθ −1 − 1) 1 + 1 − (e + e ) 2(1 − cos θ) (e − 1) (e 1 sin θ = − −i· . 2 2(1 − cos θ) = eiθ Therefore, every root z = x + iy of the original equation must therefore have the form 2 sin θ2 cos θ2 1 i θ sin θ 1 1 = − − cot , z =− +i· =− +i· 2 2 2 2 2 2(cos θ − 1) 2 1 − 2 sin2 θ2 − 1 and it follows that the real part is always x = − 1 as required.

Finally, since |AB| = 1, cos π 5 1 1 1 |DC| = k10 + |DC| = k10 + (1 − k10 ) 2 √ 2 2 1+ 5 1 . 2 The notation k10 is due to the fact that it is the length of the cord of the regular decagon, inscribed in the unit circle. 6 Find all roots of the equation z 4 + i = 0. We rewrite this equation as π z 4 = −i = exp i − + 2pπ 2 thus π π z = exp i − + p · , 8 2 It follows from π cos = 8 p ∈ Z, , p = 0, 1, 2, 3. cos π4 + 1 = 2 1+ 1 − cos π4 = 2 1− √1 2 2 √ 2+1 √ = 2 2 = 2+ 2 √ 2 , and π sin = 8 √1 2 2 √ = 2−1 √ = 2 2 2− 2 √ 2 , that 2−i 2− √ 2 , 2+i 2+ √ 2 , 1 2 2+ i z1 = 1 2 2− = −z1 = √ 1 − 2+ 2+i 2 2− √ 2 , = −i z1 = √ 1 − 2− 2−i 2 2+ √ 2 .

Finally, if a ∈ [−1, 1], then it follows from a2 − 1 = a ± i z =a± 1 − a2 , 1 − a2 ≥ 0, that |z|2 = a2 + 1 − a2 = 1, and we have proved that in this case both z1 and z2 lie on the unit circle. 3 The function f (z) = 1 2 z+ 1 z , z = 0, is also called Joukovski’s function. It was many years ago applied by Joukovski in order to describe the streamlines around the wing of an aeroplane. 4 Prove that 1 ± i are the roots of the polynomial z 4 − 2z 3 + 3z 2 − 2z + 2. Then find all its roots. First method.

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