Complex Solution Set

**mstrfrdmx** · February 5th 2011, 01:17 PM

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

**dwsmith** · February 5th 2011, 01:55 PM

Originally Posted by mstrfrdmx

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

Edit

**mstrfrdmx** · February 5th 2011, 02:07 PM

I can't see the edit though.

**Plato** · February 5th 2011, 02:19 PM

Originally Posted by mstrfrdmx

Find all complex solutions of $z^5=(z+1)^5.$

There may be a clever trick. I don't see it right off.
But this may save you some work.
$w^5-z^5=(w-z)(w^4+w^3z+w^2z^2+wz^3+z^4)$ .
Letting $w=z+1$ the first factor is 1.
Then we get a polynomial in $z^4$ .
There are four complex answers.

**Plato** · February 5th 2011, 02:33 PM

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

**mstrfrdmx** · February 5th 2011, 02:40 PM

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

**dwsmith** · February 5th 2011, 02:53 PM

Originally Posted by mstrfrdmx

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

$\displaystyle w_k=r^{1/n}\left[\cos\left(\frac{\theta+2\pi k}{n}\right)+i\sin\left(\frac{\theta+2\pi k}{n}\right)\right]$

**Plato** · February 5th 2011, 02:55 PM

Originally Posted by mstrfrdmx

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

I can help you a bit more.
$\dfrac{1}{w-1}=\dfrac{\overline{w}-1}{|w-1|^2}$

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

I still do not know where your answer comes from.

**mstrfrdmx** · February 5th 2011, 03:26 PM

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

**Unbeatable0** · February 6th 2011, 02:42 AM

Originally Posted by mstrfrdmx

Yeah, that's what I had done. The solution is $\frac{1}{w-1}$ where $w=e^{\frac{i2k\pi }{5}}, k=1, 2, 3, 4.$ 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:

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

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

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

**mstrfrdmx** · February 6th 2011, 11:15 AM

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:

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

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

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

You the coolest, doc. Muchos kudos.

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