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We just started learning about complex numbers, and I'm having some trouble on how to solve this problem. $$z^4 = (z + 1)^4$$

Was wondering if anyone could help me out.

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    $\begingroup$ What are youre thoughts on the problem so far Viktor? $\endgroup$
    – Daniel
    Jan 21 '17 at 13:19
  • $\begingroup$ I try to visualize the numbers in the complex plane, and then "turn-multiply" them, and see if they fit. I guess the real part have to be 1/2. $\endgroup$
    – Viktor
    Jan 21 '17 at 13:20
  • $\begingroup$ See math.stackexchange.com/questions/2088726/… $\endgroup$ Jan 21 '17 at 13:21
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    $\begingroup$ Set $w=1+\frac{1}{z}$, solve $w^4=1$ then solve with respect to $z$. $\endgroup$ Jan 21 '17 at 14:06
  • $\begingroup$ Get the zero on one side by subtraction and then use the conjugate rule twice: $x^2-y^2 = (x+y)(x-y)$ using the fact that $^4$ is ${^2}^2$ $\endgroup$ Jan 21 '17 at 18:50
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HINT:

There is one (real) solution that you can find by inspection:

Since $z^4 = (z+1)^4$ is kind of "symmetric" with respect to $z=-1/2$, it makes sense to evaluate this number. You will find it is a solution for the equation.

Once you find it, can you see a way of transforming this equation into a degree 3 polynomial equation? if so, and since you already know one solution, it should be easy (applying synthetic division and the quadratic formula) to find the other two roots.

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  • $\begingroup$ Is the real solution -1/2? $\endgroup$
    – Viktor
    Jan 21 '17 at 13:24
  • $\begingroup$ Indeed! it is (I just edited my answer to add this) $\endgroup$
    – Daniel
    Jan 21 '17 at 13:25
  • $\begingroup$ Expand the right side, move it over to the left like this:$$4z^3 + 6z^2 + 4z + 1 = 0$$ $\endgroup$
    – Viktor
    Jan 21 '17 at 13:32
  • $\begingroup$ Ok, very well, and can you find the other roots from there? $\endgroup$
    – Daniel
    Jan 21 '17 at 13:39
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    $\begingroup$ Yeah, polynomial division with (z + 1/2) and from there I can use the quadratic formula to find the last two/one root(s). Thanks a lot man! $\endgroup$
    – Viktor
    Jan 21 '17 at 13:43
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$(z+1)^4-z^4=((z+1)^2+z^2)((z+1)^2-z^2)=(2z^2+2z+1)(2z+1)$, which gives the answer: $$\left\{-\frac{1}{2}-\frac{1}{2}i,-\frac{1}{2}+\frac{1}{2}i,-\frac{1}{2}\right\}$$

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  • $\begingroup$ This should probably be the most easily accessible approach for someone just introduced to complex numbers.. +1 $\endgroup$ Jan 21 '17 at 18:56
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Hint: $z^4=(z+1)^4$ i.e $z=e^{2i\pi\over 4}(z+1), i=0,1,2,3$.

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Clearly $z=0$ is not solution. Thus you can divide by $z^{4}$.

You get $((z+1)/z)^4 =1$.

You know the solutions to $X^4 = 1$ in the complex numbers are $\pm 1$ and $\pm i$.

So you are down to solving $(z+1)/z = 1$, $(z+1)/z=-1$, $(z+1)/z = i$, $(z+1)/z=-i$, which is easy.

This is basically what Jack said in a comment.

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the equation implies $$ |z|^2=|z+1|^2 $$ so $$ z\bar z = z\bar z+z+\bar z +1 $$ giving $$\mathfrak{R}(z) = -\frac12$$

set $z=-\frac12 + iy$ ($y \in \mathbb{R}$)

so $$ (-\frac12 + iy)^4 = (\frac12 + iy)^4 $$ giving (since the real terms on either side cancel) $$ 4(-\frac12)^3iy + 4(-\frac12)(iy)^3 = 4(\frac12)^3iy + 4(\frac12)(iy)^3 $$ i.e. $$ y-4y^3=0 $$ so the three roots of the original cubic are $$ z= -\frac12, -\frac12 \pm \frac{i}2 $$

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The equation implies $z=t(z+1)$ with $|t|=1$ that means $z=\frac{t}{1-t}$ with $t\not=1$. Coming back to the equation we get $t^4=1$ that means $t\in\{-1,i,-i\}$ hence $z\in\{\frac{-1}{2},\frac{i}{1-i}, \frac{-i}{1+i}\}$

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