Question

How many solutions within a unit circle \(|z| < 1\) does the equation \((1 +
z)^{n + m} = z^n\) have for \(z\) complex, and \(n\), \(m\) positive integers?

Answer 1

@BSwan

Answer 2

@Kainui

Answer 3

\(\large \left(1+z\right)^m = \left(\dfrac{z}{1+z}\right)^n\)

Answer 4

i thought about it alote :-\
mmm have no clue sry , but since ur talking about Z plane
i think if u found one solution , then there is infinitly many solution (have doubt )
ill let u know if i found something else

Answer 5

updated one year ago!!!

Answer 6

After one year, no one has an answer

Answer 7

i have one constraint... tell me if i am going right re (z) >-1/2

Answer 8

please try this substitution:
\[\Large z = \rho {e^{i\theta }}\]

Answer 9

we can rewrite your equation as follows:
\[\Large 1 + z = {z^{\frac{n}{{m + n}}}}\]
and after that substitution we have:
\[\Large \rho \left( {{e^{i\frac{\pi }{2}}} + {e^{i\theta }}} \right) = {\rho ^{\frac{n}{{m + n}}}}{e^{i\frac{n}{{m + n}}\theta }}\]

Answer 10

oops..
\[\Large \rho \left( {{e^{i0}} + {e^{i\theta }}} \right) = {\rho ^{\frac{n}{{m + n}}}}{e^{i\frac{n}{{m + n}}\theta }}\]

Answer 11

Nice problem.

Answer 12

so
since |z|<1.... r^n<1

\(\Large \begin{matrix}
1+z=& z^{\frac{n}{n+m}} \\
1+r(\cos\theta+i \sin\theta ) =& r^{\frac{n}{n+m}}(\cos{\frac{n}{n+m}} \theta + i \sin{\frac{n}{n+m}} )\\
(1+r \cos \theta )+ i(r \sin \theta ) =& (r^{\frac{n}{n+m}} \cos{\frac{n}{n+m}} \theta )+ i (r^{\frac{n}{n+m}}\sin{\frac{n}{n+m}}) \\
\end{matrix} \)

Answer 13

\(\begin{matrix}
1+z=& z^{\frac{n}{n+m}} \\
1+r(\cos\theta+i \sin\theta ) =& r^{\frac{n}{n+m}}(\cos{\frac{n}{n+m}} \theta + i \sin{\frac{n}{n+m}} )\\
(1+r \cos \theta )+ i(r \sin \theta ) =& (r^{\frac{n}{n+m}} \cos{\frac{n}{n+m}} \theta )+ i (r^{\frac{n}{n+m}}\sin{\frac{n}{n+m}}) \\
\end{matrix} \)

Answer 14

so as we know imaginary part together , real part together :)

Answer 15

\( \Large \begin{matrix}

(1+r \cos \theta ) =& (r^{\frac{n}{n+m}} \cos{\frac{n}{n+m}} \theta )\text{ equation 1 } \\
(r \sin \theta ) =& (r^{\frac{n}{n+m}}\sin{\frac{n}{n+m}}\theta )\text{ equation 2 } \\
\end{matrix} \)