Complex EquationsDate: 6/14/96 at 10:44:57 From: Kees van Schaik Subject: COMPLEX-question Dear Doctor, I have a little problem: Let z be an element of the complex numbers. Then solve: (1) cos z = 2 (2) z^2 - 2z = i Date: 6/14/96 at 12:0:1 From: Doctor Anthony Subject: Re: COMPLEX-question (1) cos(z) = 2. We have cos(z) = (e^(iz)+e^(-iz))/2 = 2. So e^(iz)+e^(-iz) = 4. Multiply through by e^(iz) and we get e^(2iz) + 1 = 4e^(iz). So e^(2iz) - 4e^(iz) + 1 = 0. This is a quadratic in e^(iz) Applying the quadratic formula: e^(iz) = {4 +or- sqrt(16-4)}/2 = {4 +or- 2sqrt(3)}/2 = 2 +or- sqrt(3) iz = ln(2 +or- sqrt(3)) z = -i*ln(2 +or- sqrt(3)) = i*ln{1/(2 +or- sqrt(3))} = i*ln(2-sqrt(3)) or i*ln(2+sqrt(3)) (2) Solve z^2-2z = i We could complete the square getting z^2 - 2z + 1 = 1 + i. Then (z - 1)^2 = 1 + i and taking square roots z-1 = +or- sqrt(1+i). The square root of 1+i is more easily done if we note that 1 + i = sqrt(2)*e^(i*pi/4) and +or- sqrt(1+i) = +or- 2^(1/4)*e^(i*pi/8). And so finally z = 1 +or- 2^(1/4)*e^(i*pi/8) z = 1 +or- 2^(1/4){cos(pi/8)+i*sin(pi/8)}. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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