| Thread Closed |
Tricky complex numbers question |
Share Thread |
| Feb15-10, 01:50 PM | #1 |
|
|
Tricky complex numbers question
1. The problem statement, all variables and given/known data
Harder: given that √(−15 − 8i) = ±(1 − 4i) obtain the two solutions of the equation z² + (−3 + 2i)z + 5 − i = 0 2. Relevant equations I can easily prove √(−15 − 8i) = ±(1 − 4i) but that's not important 3. 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 |
| Feb15-10, 02:37 PM | #2 |
Recognitions:
|
You can't separate it into real and imaginary parts like that, z itself is probably complex. Just use the quadratic formula on the original equation.
|
| Thread Closed |
Similar discussions for: Tricky complex numbers question
|
||||
| Thread | Forum | Replies | ||
| Tricky complex numbers problem | Precalculus Mathematics Homework | 5 | ||
| One Question.....Complex Numbers | Calculus & Beyond Homework | 4 | ||
| Complex numbers question. | Calculus & Beyond Homework | 9 | ||
| Complex numbers question | Precalculus Mathematics Homework | 2 | ||
| tricky complex numbers proof | Calculus & Beyond Homework | 3 | ||