I'm trying to solve $z^2-2iz-3=0$, but I don't understand why my method doesn't work ! So, $$\Delta =(2i)^2-4\cdot (-3)=-4$$ and thus $$z=\frac{2i\pm 2i}{2}=0\ \ or\ \ 2i$$ but they are both no solution... what's wrong here ?
Solve $z^2-2iz-3=0$.
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complex-analysis
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1$(2i)^2-4(-3)=-4+12=8$ I think – 2017-02-17
2 Answers
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You have $(2i)^2-(4)(-3)=-4+12=8$
4
There lies some calculation mistakes in your work. We have $\Delta = (2i)^2-4(1)(-3) = 4i^2+12 =-4+12=8$.
Thus, the roots are $\displaystyle \frac{2i \pm \sqrt{8}}{2} = \pm \sqrt{2} + i$. Hope it helps.
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0Should be $\pm 4+i $ – 2017-02-17
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0Should be $\pm \sqrt{2}+i$ since you forgot to take square root of discriminant. – 2017-02-17