1. ## Complex Solution Set

Find all complex solutions of $\displaystyle z^5=(z+1)^5.$ I guess it's possible to just substitute $\displaystyle a+bi,$ collect terms, and expand, but the solution intimated a quicker way. Anyone willing to crack it receives muchos kudos from me.

2. Originally Posted by mstrfrdmx
Find all complex solutions of $\displaystyle z^5=(z+1)^5.$ I guess it's possible to just substitute $\displaystyle a+bi,$ collect terms, and expand, but the solution intimated a quicker way. Anyone willing to crack it receives muchos kudos from me.
Edit

3. I can't see the edit though.

4. Originally Posted by mstrfrdmx
Find all complex solutions of $\displaystyle z^5=(z+1)^5.$
There may be a clever trick. I don't see it right off.
But this may save you some work.
$\displaystyle w^5-z^5=(w-z)(w^4+w^3z+w^2z^2+wz^3+z^4)$.
Letting $\displaystyle w=z+1$ the first factor is 1.
Then we get a polynomial in $\displaystyle z^4$.

5. Better still just expand it.
$\displaystyle (z+1)^5=z^5+5z^4+10z^3+10z^2+5z+1$

6. Yeah, that's what I had done. The solution is $\displaystyle \frac{1}{w-1}$ where $\displaystyle w=e^{\frac{i2k\pi }{5}}, k=1, 2, 3, 4.$ I just have no idea what to do after putting everything into polar form.

7. Originally Posted by mstrfrdmx
Yeah, that's what I had done. The solution is $\displaystyle \frac{1}{w-1}$ where $\displaystyle w=e^{\frac{i2k\pi }{5}}, k=1, 2, 3, 4.$ I just have no idea what to do after putting everything into polar form.
$\displaystyle \displaystyle w_k=r^{1/n}\left[\cos\left(\frac{\theta+2\pi k}{n}\right)+i\sin\left(\frac{\theta+2\pi k}{n}\right)\right]$

8. Originally Posted by mstrfrdmx
Yeah, that's what I had done. The solution is $\displaystyle \frac{1}{w-1}$ where $\displaystyle w=e^{\frac{i2k\pi }{5}}, k=1, 2, 3, 4.$ I just have no idea what to do after putting everything into polar form.
$\displaystyle \dfrac{1}{w-1}=\dfrac{\overline{w}-1}{|w-1|^2}$

and $\displaystyle \overline{e^{u}}=e^{\overline{u}}$.

9. I may just be slow, but I don't really see how the nth roots formula helps in this case.

10. Originally Posted by mstrfrdmx
Yeah, that's what I had done. The solution is $\displaystyle \frac{1}{w-1}$ where $\displaystyle w=e^{\frac{i2k\pi }{5}}, k=1, 2, 3, 4.$ I just have no idea what to do after putting everything into polar form.
I'm not sure if you meant that you got this answer or you just know it's the answer and need to get to it.

If you want to know how to get to it, this may help:

$\displaystyle \displaystyle{z^5=(1+z)^5 \Leftrightarrow (1+z^{-1})^5=1}$

which means all the solutions are $\displaystyle 1+z^{-1}=w_k$ where the $\displaystyle w_k$

are the roots of unity of order 5. Equivalently, $\displaystyle z=\frac{1}{w_k - 1}$

11. Originally Posted by Unbeatable0
I'm not sure if you meant that you got this answer or you just know it's the answer and need to get to it.

If you want to know how to get to it, this may help:

$\displaystyle \displaystyle{z^5=(1+z)^5 \Leftrightarrow (1+z^{-1})^5=1}$

which means all the solutions are $\displaystyle 1+z^{-1}=w_k$ where the $\displaystyle w_k$

are the roots of unity of order 5. Equivalently, $\displaystyle z=\frac{1}{w_k - 1}$
You the coolest, doc. Muchos kudos.