given: $z^2 + 0.8 \sqrt2 z + 0.64 = 0 $

Then, I am using the quadratic equation:

$ z_{1,2} = \frac{-0.8 \sqrt2 \pm \sqrt{(0.8\sqrt2)^2-4 \cdot 0.64}}{2} $

Wolfram Alpha says it the end there should be: $-0.4 \sqrt2 \pm j 0.4 \sqrt2$, but I get something different:

$$ = \frac{-0.8\sqrt2\pm0.8\sqrt2-2\cdot0.8j}{2} $$ $$ = -0.4\sqrt2\pm0.4\sqrt2-1.6j$$


1 Answer 1


You seem to be computing, on the square root side, something like, $a$ and $b$ being positive, $\sqrt{a-b}=\sqrt{a}-j\sqrt{b}$, which is wrong. The square root is not linear.

Here, the discriminant $\Delta = (0.8\sqrt{2})^2-4\times 0.8^2 = -2\times 0.8^2$ is negative. Hence, the roots are:

$$ z_r= \frac{-0.8\sqrt{2}\pm j\sqrt{-\Delta}}{2} = 0.4\sqrt{2}(-1 \pm j)$$

agreeing with Wolfram Alpha.


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