2

I have to solve $$z^2-(1+i)z-i=0,$$ then I get $$\Delta =8i=8e^{i\pi/2}.$$

The solution are $z_{1,2}=(1+i)\pm \sqrt{\Delta }$, but for $\sqrt \Delta$, I have $$\sqrt\Delta =2\sqrt 2 e^{i(\frac{\pi}{4}+k\pi)}$$ with $k=0,1$. So which square root I have to take, $2\sqrt 2 e^{i\frac{\pi}{4}}$ or $2\sqrt 2 e^{i\frac{5\pi}{4}}$ ?

idm
  • 11,824

1 Answers1

3

Short answer: Both.

The $i$ is only a notation and $\sqrt{8i}$ is just a symbol that does not have to be well defined. Similarly you could ask yourself what is the root of $x^2-1$, do you have to take $+1$ or $-1$? There is another nice example in the complex numbers: What is the value of $i^i$? You can show that it must be a real, but there are infinite possible values. See this question here.

Conventionally we define the function $x \mapsto \sqrt{x}$ to return a nonnegative number, if $x$ is real and nonnegative. We cannot do that as simply in the complex numbers, there we have to restrict the domain or co-domain in order to get an unique answer. In your case we can restrict the co-domain to $\{re^{i \phi} | r\geq 0 , -\pi < \phi \leq \pi \}$ in order to get an unique value.

Community
  • 1
flawr
  • 16,533
  • 5
  • 41
  • 66
  • In that application one does not have to consider both square-roots, as there is a $\pm$ in the formula. It is sufficient to consider one and which one does not matter. Moreover, what you write does not restrict the co-domain at all. – quid Feb 13 '15 at 11:34
  • 1
    Usually, "symbols" very much "have to be well defined". In the present case "$\sqrt{8i}$ is just a symbol that" should not be used at all. – Did Feb 13 '15 at 12:06