Solve $z^5 - iz^3+iz^2+1 =0$

Question

Solve for $z$: $$z^5 - iz^3+iz^2+1 =0$$ I've ruled out the approach with euler's representation, de moivre's way led me to $\cos4\phi + \sin3\phi-\sin2\phi +1 = 0$ (with the assumption that the length of $z$ is $1$, which I derived from the fact that $i$ is one of the solutions, which shows trivially through horner's algorithm). Substituting $z$ with $a+ bi$ got me some very ugly numerical solutions for $a$, and as previously mentioned, $i$ is a solution that leads to the equation $$z^4+iz^3+(-1-i)z^2+z+i=0$$I need help.

score 2 · Accepted Answer · answered Oct 20 '18 at 15:09

2

$$z^5 - iz^3+iz^2+1 =z^3(z^2-i)+i(z^2-i)=(z^3+i)(z^2-i)$$

You should be able to solve it now.

answered Oct 20 '18 at 15:09

edm

5,640

wow, thanks, may I ask how you figured this out, was there a methodical approach involved or just mathematical intuition? – uros kozole Oct 21 '18 at 20:03
@uroskozole I simply guess that this factorisation may work. You may say it is intuition. – edm Oct 22 '18 at 09:11

score 1 · Answer 2 · answered Oct 20 '18 at 15:08

1

Hint:

$$z^3(z^2-i)+i(z^2-i)=?$$

answered Oct 20 '18 at 15:08

lab bhattacharjee

274,582

Solve $z^5 - iz^3+iz^2+1 =0$

2 Answers2