$(z*) * z + 4iz + 4i = 0$
How to solve an equation like this? Substituting z with a + bi gives:
$z^2 + 4i(a+bi)+4i = 0$
$z^2+ 4ai - 4b + 4i = 0$
What to do next?
(z*) * z + 4iz + 4i = 0, complex equation
0
$\begingroup$
complex-numbers
quadratics
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0do you mean $$z^2+4iz+4i=0$$? – 2017-01-16
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0I think it you can write like that too yes – 2017-01-16
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2What does the $z*$ inside the parentheses mean? And what does the $*$ outside the parentheses mean? – 2017-01-16
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0Does $(z*)*z$ mean $\,\overline{\overline{z}} \cdot z\,$, or $\,\overline{z} \cdot z\,$ or ..? – 2017-01-16
1 Answers
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solving the equation $$z^2+4iz+4i=0$$ we get $$z_1=-2i+2\sqrt{-1-i}$$ or $$z_2=-2i-2\sqrt{-1-i}$$ we get this using the formula for a quadratic equation of the form $$x^2+px+q=0$$ which has the Solutions $$x_1=-\frac{p}{2}+\sqrt{\frac{p^2}{4}-q}$$ or $$x_2=-\frac{p}{2}-\sqrt{\frac{p^2}{4}-q}$$
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0Perhaps you should mention **how** you did arrive at those results: quadratic equation's roots' formula – 2017-01-16
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0i have wrote this with the formula, thank you for the hint – 2017-01-16
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0you are welcome! – 2017-01-16
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0Is there a way of writing $z=-2i±2i*sqrt(i+1)$ as $z = a + bi$ – 2017-01-16
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0yes there is a way, where should i write this? – 2017-01-16