0

$z^4 + 1 =0$, looking at a relatively similar question I concluded that the next step is to factorise into $(z^2 + i) (z^2 - i) = 0$. However, I'm not sure what the next step should be? Do i continue to factorise?

Adam Jones
  • 105
  • Sure, that would work. You should end up with four complex numbers, which you can take the fourth power of to make sure that they are the correct solutions. – Bilbottom May 10 '18 at 11:46
  • 1
    Yes you can ! Or also simply use the fact that $$z^4=-1=e^{i\pi(2k+1)}\implies z=e^{\frac{i\pi(2k+1)}{4}}, k=0,1,2,3$$ – Surb May 10 '18 at 11:47
  • The duplicate is not an exact duplicate. It involves the same method, but isnt the same question. – Rhys Hughes May 10 '18 at 12:00

1 Answers1

2

$$z^4=-1+0i$$ $$\to z^4=\cos[(2k+1)\pi]+i\sin[(2k+1)\pi]$$ $$\to z= \cos\bigg[\frac{(2k+1)\pi}{4}\bigg]+i\sin\bigg[\frac{(2k+1)\pi}{4}\bigg]$$

$$k \in \Bbb Z$$

And go from there, by inserting $k=0,1,-1$ etc

Rhys Hughes
  • 12,842