2

I have here an complex equation in $\mathbb{C}$ numbers: $$z^4 + 1 = \dfrac{1 - iz^3}{1 - z^4}.$$

My question is how to solve this complex equation? What is the best way to solve it?

My attempt: Multiply both sides by $1-z^4$ to get $$(z^4+1)(1-z^4)=1-iz^3$$ so $1-z^8=1-iz^3$ gives $z^8=z^3i$. How to continue?

ə̷̶̸͇̘̜́̍͗̂̄︣͟
  • 27,005
Andrej
  • 153
  • 2
    What have you tried so far? Hint: cross multiply by the denominator. – Blitzer Jun 02 '21 at 08:18
  • 1
    $(z^4+1)(1-z^4)=-(z^4-1)(z^4+1)=-(z^8-1)=1-z^8$ – Alessandro Jun 02 '21 at 08:18
  • Yes, I came through my step to here. I got the equation $$z^8 = z^3i$$ and then I stopped, cause I don't know how to get a solutions from here. Can anybody help me? – Andrej Jun 02 '21 at 08:21
  • 3
    Please show your efforts by editing your attempt into the post. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Jun 02 '21 at 08:23
  • @Andrej Guess you are supposed to know how to solve equations like $z^n = i$. If not, what is the context in which you found this problem? – dxiv Jun 02 '21 at 08:24
  • 1
    One answer is $z=0$. If $z \ne 0$, then divide both sides by $z^3$. – Steven Alexis Gregory Jun 02 '21 at 08:25
  • Ok, my attempt. $$z^4 + 1 = \dfrac{1 - iz^3}{1 - z^4}$$. First multiply equation by $1 - z^4$. $$(z^4 + 1)(1 - z^4) = 1 - iz^3$$. Then I get the equation $z^8 = z^3i$ and then I stop. – Andrej Jun 02 '21 at 08:26
  • 1
    $z=0$ solves the equation. So assume $z\not=0$ and obtain $z^5=i$. You should be able to solve the equation. –  Jun 02 '21 at 08:29
  • Ok, thanks for your help. I think will now solve the equation. – Andrej Jun 02 '21 at 08:30
  • 1
    I think the result would be more recognisable if you make the substitution $z=iw$, so that $z^8=z^3i$ becomes $w^8=w^3$. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Jun 02 '21 at 08:31
  • 1
    Ok, either way. For equation $w^8 = w^3$ and rearrange the equation on $w^3(w^5 - 1) = 0$. I got the solutions: $w_{1,2,3} = 0$, $w_4 = \cos(0) + i \sin(0) = 1$, $w_5 = \cos(\dfrac{2\pi}{5}) +i \sin(\dfrac{2\pi}{5})$, $w_6 = \cos(\dfrac{4\pi}{5}) + i \sin(\dfrac{4\pi}{5})$, $w_7 = \cos(\dfrac{6\pi}{5}) + i \sin(\dfrac{6\pi}{5})$ and $w_8 = \cos(\dfrac{8\pi}{5}) + i \sin(\dfrac{8\pi}{5})$. – Andrej Jun 02 '21 at 09:00
  • 1
    For equation $z^5 = i$ I got the solutions: $z_0 = \cos(\dfrac{\pi}{10}) + i \sin(\dfrac{\pi}{10})$, $z_1 = \cos(\dfrac{\pi}{2}) + i \sin(\dfrac{\pi}{2}) = i$, $z_2 = \cos(\dfrac{9\pi}{10}) + i \sin(\dfrac{9\pi}{10})$, $z_3 = \cos(\dfrac{13\pi}{10}) + i \sin(\dfrac{13\pi}{10})$ and $z_4 = \cos(\dfrac{17\pi}{10}) + i \sin(\dfrac{17\pi}{10})$. – Andrej Jun 02 '21 at 09:06
  • Is this right solution? Anyone? – Andrej Jun 02 '21 at 09:06
  • 2
    @Andrej looks correct to me. – Klaus Jun 02 '21 at 09:14

0 Answers0