Calculus Analyse lc-4, Complex Numbers Examples by Mejlbro L.
By Mejlbro L.
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All things considered we get the solutions ±1 ± i with all four possible combinations of the signs. Third variant. Write the equation in the form z 4 + 4 = 0. Then by a rearrangement, 0 = z 4 + 4 = (z 4 + 4z 2 + 4) − 4z 2 = (z 2 + 2)2 − (2z)2 = (z 2 + 2z + 2)(z 2 − 2z + 2) = {(z + 1)2 + 1}{(z − 1)2 + 1}, from which we get the solutions −1 ± i og 1 ± i. √ 2) From z 3 = 1 + i = ( 2) π4 +2pπ we get √ 6 π z = ( 2) 12 , p = 0, 1, 2. + 2ππ 3 Here, √ 6 π ( 2) 12 = = = = √ 6 π π + i sin 2 cos 12 12 ⎫ ⎧ π π⎪ ⎪ ⎪ ⎪ ⎬ ⎨ 1 + cos 1 − cos √ 6 6 +i 6 2 ⎪ ⎪ 2 2 ⎪ ⎪ ⎭ ⎩ ⎫ ⎧ √ √ ⎬ ⎨ 1 √ 3 1 3 6 1+ +i 1− 2 ⎩ 2 2 2 2 ⎭ √ 6 2 √ 1 (4 + 2 3) + i 8 √ 1 (4 − 2 3) 8 √ 6 √ √ 3 √ ( 3 + 1)2 + i ( 3 − 1)2 = 2 2 √ √ 1 √ = 3 + 1 + i( 3 − 1) .
D. Just calculate; apply, if necessary the complex definitions of the trigonometric functions. I. 1) ei π 2 π π + i sin = i. 2 2 = 3e{cos π + i sin π} = −3e. = cos 2) 3e1+iπ 3) From ei π 2 = i we get π π 1 − ei 2 π 1−i iπ 4 − e−i 4 = + 2i sin π + e i 1+i 4 1+e 2 = √ √ −i(1 + i) + i 2 = i( 2 − 1). 1+i 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.
2) Write the polynomial P (x) = x4 +16 as a product of polynomials of second degree of real coefficients. A. Binomial equation and factorization of a polynomial. D. Find the complex roots, and use the solutions it the factorization of P (x). Alternatively, add a term and subtract it again. I. 1) We write the binomial equation in the form z 2 = −16 = (24 )π+2pπ , from which z = (2) π4 +p π2 , Since (2) π4 = √ p = 0, 1, 2, 3. √ √ 2 + i 2 and (1) π2 = i, the solutions are given by √ 2 + i 2, √ √ − 2 + i 2, √ √ − 2 − i 2, √ √ 2 − i 2.