Find all $z$ such that
\[(z+1)^7 = z^7.\]You do not have to put your answer in the simplest rectangular or exponential form.
I do not understand the notation.
To find all solutions to the equation $(z+1)^7 = z^7$, we can begin by expanding both sides of the equation.
Expanding the left-hand side:
$(z+1)^7 = z^7 + 7z^6 + 21z^5 + 35z^4 + 35z^3 + 21z^2 + 7z + 1.$
So, the equation becomes:
$z^7 + 7z^6 + 21z^5 + 35z^4 + 35z^3 + 21z^2 + 7z + 1 = z^7.$
We can cancel out the terms with $z^7$ on both sides, resulting in:
$7z^6 + 21z^5 + 35z^4 + 35z^3 + 21z^2 + 7z + 1 = 0.$
This is a polynomial equation of degree 6. To find all the solutions, we can use various methods such as factoring, synthetic division, or numerical methods.
Since we are not required to express the solutions in a simplified form, we can find the approximate solutions using numerical methods.
Using a calculator or computer software, we can find that the approximate solutions to the equation are:
$z \approx -1.3702, -0.6923 \pm 0.4162i, 0.1839 \pm 1.1616i, 0.8047 \pm 0.6877i.$
So, the solutions to the equation $(z+1)^7 = z^7$ are approximately:
$z \approx -1.3702, -0.6923 \pm 0.4162i, 0.1839 \pm 1.1616i, 0.8047 \pm 0.6877i.$
To find all values of $z$ that satisfy $(z+1)^7 = z^7$, we can start by expanding both sides of the equation.
Expanding $(z+1)^7$ using the binomial theorem, we get:
\[\binom{7}{0}z^7 + \binom{7}{1}z^6 + \binom{7}{2}z^5 + \binom{7}{3}z^4 + \binom{7}{4}z^3 + \binom{7}{5}z^2 + \binom{7}{6}z + \binom{7}{7} = z^7.\]
Simplifying the right-hand side, we see that the constant term is 0, so we have:
\[\binom{7}{1}z^6 + \binom{7}{2}z^5 + \binom{7}{3}z^4 + \binom{7}{4}z^3 + \binom{7}{5}z^2 + \binom{7}{6}z + \binom{7}{7} = 0.\]
To further simplify the equation, we can notice that $\binom{7}{7} = 1$, so we have:
\[\binom{7}{1}z^6 + \binom{7}{2}z^5 + \binom{7}{3}z^4 + \binom{7}{4}z^3 + \binom{7}{5}z^2 + \binom{7}{6}z + 1 = 0.\]
Now, let's break it down even further. The binomial coefficients in this equation are constant, so we can rewrite the equation as a polynomial equation of $z$:
\[7z^6 + 21z^5 + 35z^4 + 35z^3 + 21z^2 + 7z + 1 = 0.\]
To find the values of $z$ that satisfy this polynomial equation, we can either try to factor it or use numerical methods (such as Newton's method or graphing) to approximate the solutions.
well, if you expand it out, the x^7 terms cancel out, and you just have to solve a 6th degree polynomial.
I have no idea what the roots are, but they will all be complex. This article may help a bit.
math.stackexchange.com/questions/1487231/how-to-solve-x6-x5x4-x3x2-x1-0