Solving Complex Variables EquationsDate: 09/01/98 at 08:12:41 From: derek zumpolle Subject: Complex numbers: determining z Determine all z such that: 2z + (conjugate of z)^2 = -1+6i 4z^2 - (conjugate of z)^2 = 20i (conjugate of z) z^3 = 4i Date: 09/01/98 at 08:48:38 From: Doctor Jerry Subject: Re: Complex numbers: determining z Hi Derek, If z = x+i*y, then the first equation is: 2(x+i*y) + (x-i*y)^2 = -1+6i 2x + 2*i*y + x^2 - y^2 - 2i*x*y = -1+6i So: 2x + x^2 - y^2 = -1 2y - 2x*y = 6 Solve this system simultaneously for x and y. The others can be done in the same way. - Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 09/01/98 at 11:51:39 From: Doctor Anthony Subject: Re: complex numbers: determining z I shall use z* to represent the conjugate of z. To find the z such that 2z + (z*)^2 = -1+6i: 2z + (z*)^2 = -1+6i 2(x+iy) + [x^2 - i 2xy - y^2] = -1 + 6i x^2 + 2x - y^2 + i[2y - 2xy] = -1 + 6i x^2 + 2x - y^2 = -1 and 2y(1-x) = 6 So: y = 3/(1-x) Substituting for y: x^2 + 2x - 9/(1-x)^2 = -1 (x^2 + 2x + 1)(1-x)^2 = 9 (x+1)^2 (x-1)^2 = 9 [x^2 - 1]^2 = 9 Taking square roots: x^2 - 1 = +/-3 So x^2 = -2 (which is not possible as x is real) or x^2 = 4. Thus x = +/- 2. Then if x = 2, y = 3/(1-2) = -3 and if x = -2, y = 3/(1+2) = 1 So two solutions are z = 2 - 3i and z = -2 + i. For 4z^2 - (z*)^2 = 20i, factorize the lefthand side: (2z-z*)(2z+z*) = 20i Since: 2z = 2x + 2iy z* = x - iy we have: 2z - z* = x + 3iy 2z + z* = 3x + iy Remultiplying the brackets: 3x^2 + 10ixy - 3y^2 = 20i Therefore: 3x^2 - 3y^2 = 0 y^2 = x^2 y = +/-x and: 10xy = 20 xy = 2 Thus: x = sqrt(2), y = sqrt(2) or x = -sqrt(2), y = -sqrt(2) Then the two solutions are z = sqrt(2)[1 + i] and z = -sqrt(2)[1 + i]. For (z*) z^3 = 4i: (x - iy)^3 = 4i x^3 + 3x^2 (-iy) + 3x(-iy)^2 + (-iy)^3 = 4i x^3 - 3xy^2 + iy[y^2 - 3x^2] = 4i So x(x^2 - 3y^2) = 0 and y[y^2 - 3x^2] = 4 Thus x = 0 and y^3 = 4, which implies y = 4^(1/3), or x^2 = 3y^2, which implies: y[y^2-9y^2] = 4 -8y^3 = 4 y^3 = -1/2 y = (-1/2)^(1/3) Thus: x^2 = 3(-1/2)^(2/3) x = sqrt[3(-1/2)^(2/3)] So the two solutions are: z = 0 + i 4^(1/3) z = sqrt[3(-1/2)^(2/3)] + i (-1/2)^(1/3) - Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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