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MÄLARDALENS HÖGSKOLA School of education, culture and communication Erik Darpö MAA150 Vektoralgebra Autumn term 2014 This text is a translated excerpt from a compendium written by M. Herschend and myself for a course at Uppsala University in 2007. Equations with complex coefficients We will study two types of equations: quadratic equations, which are of the form az 2 + bz + c = 0, with a, b, c ∈ C, and binomial equations: z n = w, where w is a fixed complex number and n a positive integer. Quadratic equations Using complex numbers all quadratic equations can be solved, including those with complex coefficients. We start by solving equations of the type z2 = w where w ∈ C is a given complex number. For example: z 2 = 3 − 4i. (1) Setting z = x + iy gives z 2 = (x + iy)2 = x2 + 2xyi − y 2 . Equation (1) can thus be written as x2 − y 2 + 2xyi = 3 − 4i. Identifying the real and imaginary parts of the two sides, we get the following system of equations: 2 x − y 2 = 3, (2) 2xy = −4. Solving the second equation for y gives 2 y=− , x (3) which, inserted into the first equation produces the following equation: 4 =3 x2 4 2 ⇔ x − 3x − 4 = 0 r r 3 9 3 25 3 5 ⇔ x2 = ± +4= ± = ± . 2 4 2 4 2 2 x2 − Since x2 > 0, this means that x2 = 3 5 8 + = =4 2 2 2 1 and thus x = ±2. Using the equation (3), we can now find the values of y. If x = 2, then 2 2 y = − = − = −1 x 2 whilst x = −2 gives −2 y=− = 1. 2 Hence, the solutions to Equation (1) is z = x + iy = ±(2 − i). An alternative way to solve the equation is to use the identity |z|2 = x2 + y 2 . If z is a solution to (1) then √ √ |z|2 = |z 2 | = |3 + 4i| = 9 + 16 = 25 = 5. This means that in addition to the system (2) we get a “bonus equation” x2 +y 2 = 5. Altogether: 2 x − y 2 = 3, x2 + y 2 = 5, 2xy = −4. Adding the two first of these equations to each other, we get 2x2 = 8. Hence x2 = 4 and x = ±2. As before, we use Equation (3) to find y, and get again the solutions z = x + iy = ±(2 − i). Using this method, any equation of the form z 2 = w can be solved. Combined with the techinque of completing the square, it allows us to tackle general quadratics with real or complex coefficients. Example: Solve the equation z 2 + (2 − 4i)z + 2 − 16i = 0. Solution: We start by completing the square: z 2 + (2 − 4i)z + 2 − 16i = 0 2 2 2 − 4i 2 − 4i − + 2 − 16i = 0 ⇔ z+ 2 2 ⇔ (z + 1 − 2i)2 = (1 − 2i)2 − 2 + 16i = −3 − 4i − 2 + 16 = −5 + 12i, that is, (z + 1 − 2i)2 = −5 + 12i. Setting w = z + 1 − 2i, we have w2 = −5 + 12i. This equation can be solved (for w) as in the previous example. We set w = x + iy and obtain x2 − y 2 + 2xyi = −5 + 12i Identifying the real and imaginary parts in this equation yields the system 2 x − y 2 = −5, 2xy = 12. As before, we utilise the fact that x2 + y 2 = |w|2 = |w2 | = | − 5 + 12i| = 2 √ 25 + 144 = √ 169 = 13 to get 2 x − y2 x2 + y 2 2xy = −5, = 13, = 12. Adding the first two equations gives 2x2 = −5 + 13 = 8, that is, x2 = 4 and thus x = ±2. From the equation 2xy = 12 follows that y= 6 . x If x = 2 then y = 3, whilst x = −2 gives y = −3. Consequently, w = x + yi = ±(2 + 3i). Finally, we insert these solutions into the identity w = z + 1 − 2i, and solve for z. This gives z = −1 + 2i + w = −1 + 2i ± (2 + 3i), and so z1 = −1 + 2i + 2 + 3i = 1 + 5i, z2 = −1 + 2i − (2 + 3i) = −3 − i. are the solutions to the equation z 2 + (2 − 4i)z + 2 − 16i = 0. The binomial equation We shall now consider equations of the form zn = w where w ∈ C is a given complex number and n a positive integer. Example: Solve the equation z 4 = −4. Solution: The strategy here is to write both sides of the equation on polar form. Since z = r(cos θ + i sin θ), where r = |z| and θ = arg(z), we have z 4 = r4 (cos θ + i sin θ)n = r4 (cos 4θ + i sin 4θ). The right hand side, on polar form, is −4 = 4(cos π + i sin π). The equation z 4 = −4 now becomes r4 (cos 4θ + i sin 4θ) = 4(cos π + i sin π) giving rise to the system r4 4θ = 4, = π + k2π, k ∈ Z. 3 √ 1 Since r > 0 it follows that r = 4 4 = 2. The argument θ of z is determined using the second equation: π π θ = θk = + k , k ∈ Z. 4 2 Hence we have the solutions √ z = zk = 2(cos θk + i sin θk ). Sine and cosine are periodic functions, with period 2π. Therefore, zk = zk+4 for all k ∈ Z. This means that the equation has four distinct solutions √ π 1 π √ 1 = 2 √ + √ i = 1 + i, z0 = 2 cos + i sin 4 4 2 2 √ √ 3π 1 3π 1 z1 = 2 cos + i sin = 2 − √ + √ i = −1 + i, 4 4 2 2 √ √ 5π 1 5π 1 z2 = 2 cos + i sin = 2 − √ − √ i = −1 − i, 4 4 2 2 √ √ 7π 1 7π 1 z3 = 2 cos + i sin = 2 √ − √ i = 1 − i. 4 4 2 2 The solutions may be viewed in the Argand plane as the corners of a square with side 2, centered in the origin. The method used above can be generalised to solve arbitrary equations on the form zn = w where w ∈ C is a fixed number different from 0, and n > 1. First, write z on polar form: z = r(cos θ + i sin θ) with r = |z| and θ = arg(z). Then z n = rn (cos nθ + i sin nθ). Now write the right hand side on polar form: w = |w|(cos α + i sin α). n The equation z = w becomes rn (cos nθ + i sin nθ) = |w|(cos α + i sin α), having the solutions r θ 1 = |w| n , = nπ + k 2π n , k ∈ Z. So the numbers z that satisfy the equation z n = w are 1 π 2π π 2π z = zk = |w| n cos +k + i sin +k , k∈Z n n n n Again, due to the periodicity of the trigonometic functions sin and cos, zn = z0 , zn+1 = z1 , etcetera. Hence we only need to count the solutions z0 , z1 , . . . , zn−1 , since every other solution zk is equal to one of these. The solutions may also be expressed using the complex exponential function. The numbers zk are then written as z = |w| n ei( n +k 1 π 2π n ) , k = 0, 1, . . . n − 1. They constitute the corners of a regular n-gon in the Argand plane, centered in 0. 4