Given,
\(z^{10} + 31z^5 -32 = 0\)
Let \(k = z^5\). Then,
\(k^2 + 31k -32 = 0\)
\((k - 1) (k + 32) = 0\)
\(k \in \{-32,1\}\)
\(z^5\in \{ -32, 1\}\)
Consider,
\(z^5= -32\)
We see that \(1 = 1^r = (e^{i2\pi})^r = e^{i2\pi r}\) (Recall Euler's form, 1 can be a complex number with argument 2π) So the RHS of the equation can be taken as,
\(z^5= -32e^{i2\pi r}\)
\(z = (-32e^{i2\pi r})^{\frac 15}\)
\(z = -2e^{i\frac {2\pi r}5}\)
This represents the 5 complex roots of the equation \(z^5= -32\) where r is an integer such that 0 ≤ r ≤ 4, or,
\(r \in \{0,1,2,3,4\}\)
\(\rightarrow\frac{2\pi r}5 \in \left\{0,\frac{2\pi}5, \frac{4\pi}5, \frac{6\pi}5,\frac{8\pi}5\right\} \quad .....(1)\)
[Note:- The n roots of the equation \(x^n = a\) are in the form \(x =a^\frac 1n e^{i\frac {2\pi r}n}\) where r is an integer variable such that p ≤ r ≤ q where q - p + 1 = n. Usually r is taken 0 ≤ r ≤ n - 1.]
Taking z in polar form,
Hence there are 5 solutions.
