- #1
thomas49th
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Homework Statement
Harder: given that
√(−15 − 8i) = ±(1 − 4i) obtain the two solutions of the equation
z² + (−3 + 2i)z + 5 − i = 0
Homework Equations
I can easily prove √(−15 − 8i) = ±(1 − 4i) but that's not important
The Attempt at a Solution
I would of thought that a compex solution would be a + b and a - b, but a quick glance at the answers shows 2 completely different complex numbers - no complex conjugates.
Well seperating the equation into real and imaginary parts then solving for z:
real:
(z² - 3z + 5) = 0
=> [tex] z = \frac{3\pm i \sqrt{11}}{2}[/tex]
imag:
(2z - 1) = 0
=> z = 0.5
This isn't taking me anywhere nice...
Ideas!? :)
Thanks