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**Sample text**

Com 47 Complex Funktions c-1 The binomial equation and then √ 10 − 2 5 4 π sin = 5 is found as above. B 1 k_10 A k_10 D E C k_10 1-k_10 Alternatively, the example can be solved geometrically by noting that as BCD. Then |AB| |BC| = , |BC| |CD| ABC is the same angle 1 k10 = . k10 1 − k10 thus We obtain the equation of second degree 2 k10 + k10 − 1 = 0, hence k10 1 + = − (−) 2 1 +1= 4 √ 5−1 , 2 where we have exploited that k10 > 0. Finally, since |AB| = 1, cos π 5 1 1 1 |DC| = k10 + |DC| = k10 + (1 − k10 ) 2 √ 2 2 1+ 5 1 .

Throw in continuous professional development, and you get truly interesting work, in a genuinely inspirational business. uk Applicants must be British citizens. GCHQ values diversity and welcomes applicants from all sections of the community. We want our workforce to reflect the diversity of our work. 4 It is worth mentioning that a division by z − {1 − i} = z − 1 + i will give some very unpleasant calculations and that such a division only reduces the problem to a messy equation of third degree.

If we subtract twice times the latter polynomial of (5), from this polynomial, we obtain the following reduce polynomial, which also has the wanted roots as some of its roots, (10+44i)z 3 +(−189−150i)z 2 +(388−128i)z+(−148+96i). Since 10 + 44i = 2(5 + 22i), we obtain a “nicer” expression which we multiply this polynomial by 5 − 22i, hence 1018z 3 +(−4245+3408i)z 2 +(−876−9176i)z+(1372+3736i). We have now reduced our system to ⎧ ⎨ 20z 4 +(32+64i)z 3 +(−81+96i)z 2 +(−58−88i)z+(44+12i), (6) ⎩ 1018z 3 +(−4245+3408)z 2 +(−876−9176i)z+(1372+3736i).