I have this equation to solve: $$z^4-z*(1-i)^8=0$$ I was thinking that if I can factor z and then divide both sides by it I can get a simpler equation, convert them to trigonometric form and solve it,but I'm not sure I can do that since it is not said whether z can't be 0. How can I approach this differently? Thank you in advance!

  • $\begingroup$ $1-i$ is easy to raise to the 8th power. $\endgroup$
    – B. Goddard
    Commented Oct 25, 2021 at 17:56
  • $\begingroup$ Assume the 8 power converted to 3 power and than factor. $\endgroup$
    – Moti
    Commented Oct 25, 2021 at 18:05
  • $\begingroup$ This is just the equation $z^3=(1-i)^8=16$ and the trivial solution $z=0$. $\endgroup$ Commented Oct 25, 2021 at 18:13
  • $\begingroup$ $(1-i)^2=-2i$ , $(1-i)^4=(-2i)^2=-4$ $\endgroup$ Commented Oct 26, 2021 at 3:32

2 Answers 2


At first, you can gain the trivial solution, $z=0$, and another solution is solution of $z^3=(1-i)^8$.

And you must change $(1-i)^8$ to polar form, for the conveinience of calculation. $$ (1-i)^8={\sqrt2}^8\left(\frac1{\sqrt2}-\frac1{\sqrt2}i\right)^8={\sqrt2}^8\left(\cos\frac{-\pi}4+i\sin\frac{-\pi}4\right)^8=16(e^{-\frac{\pi i}4})^8=16e^{-2\pi i}=16 $$

Then your solution is complex cube root of 16. And these are easily calculated.

  • $\begingroup$ I am not sure your conversion is right. $\endgroup$
    – Moti
    Commented Oct 25, 2021 at 18:13
  • $\begingroup$ @Moti It is right. wolframalpha.com/input/?i=%281-i%29%5E8 $\endgroup$
    – MH.Lee
    Commented Oct 25, 2021 at 18:15
  • $\begingroup$ @Moti $(1-i)^8=(1-2i+i^2)^4=(1-2i-1)^4=(-2i)^4=16$ $\endgroup$
    – Andrei
    Commented Oct 25, 2021 at 18:19
  • $\begingroup$ I tried it in a different way and got a result that is less clean, so I will need to see why the difference. In any case - a solution is always possible with factorization by just assuming that the value is a power of 3. BTW, you claim that there are three different values with power 3 that may be used - I wonder if all result the same 4 solutions. $\endgroup$
    – Moti
    Commented Oct 25, 2021 at 18:26

All that is mechanized in CASes,e.g. by the command of Maple

solve(z^4 - z*(1 - I)^8 = 0);

$$0,2 \,2^{\frac{1}{3}},-2^{\frac{1}{3}}+\mathrm{I} \sqrt{3}\, 2^{\frac{1}{3}},-2^{\frac{1}{3}}-\mathrm{I} \sqrt{3}\, 2^{\frac{1}{3}} $$


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