Real Functions in One Variable Complex Numbers Examples by Leif Mejlbro

By Leif Mejlbro

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D. Calculate |z1 + z2 |2 and estimate. I. First calculate |z1 + z2 |2 = (z 1 + z 2 )(z1 + z2 ) = |z1 |2 + |z2 |2 + z 1 z2 + z 2 z1 = |z1 |2 + |z2 |2 + 2 Re(z 1 z2 ). e. 2 2 (|z1 | − |z2 |) ≤ |z1 + z2 |2 ≤ (|z1 | + |z2 |) . Finally, by taking the square root, ||z1 | − |z2 || ≤ |z1 + z2 | ≤ |z1 | + |z2 |. Trust and responsibility Please click the advert NNE and Pharmaplan have joined forces to create NNE Pharmaplan, the world’s leading engineering and consultancy company focused entirely on the pharma and biotech industries.

E. 3 + 4i = ±(2 + i). Then by insertion −3 − i, 1+i z = −1 ± (2 + i) = C. The solutions of normed equations of second degree are checked by using that the sum of the roots is the coefficient of z with opposite sign, and the product of the roots is equal to the constant term. In the present case we get (−3 − i) + (1 + i) = −2, (−3 − i)(1 + i) = −3 + 1 + i(−3 − 1) = −2 − 4i. D. 3 Find in the form z = a + ib, a, b ∈ R, the solutions of the equation z 2 − (5 + 5i)z + 13i = 0. A. Complex equation of second degree.

I. When we multiply by −i we obtain the equivalent normed equation (2) z 2 − (3 − 2i)z + 5 − i = 0. Then by the usual solution formula known from high school, z = = = = = = 1 2 1 2 1 2 1 2 1 2 3 − 2i ± (3 − 2i)2 − 4(5 − i) 3 − 2i ± √ 9 − 4 − 12i − 20 + 4i 3 − 2i ± √ −15 − 8i 3 − 2i ± (4i)2 + 1 − 2 · 4i · 1 3 − 2i ± (1 − 4i)2 = 1 {3 − 2i ± (1 − 4i)} 2 2 − 3i, 1 + i. C. The sum of the roots is (2 − 3i) + (1 + i) = 3 − 2i, which is the coefficient of z of the opposite sign in the normed equation (2).

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