𝑧^5 = −32

asked by guest
on Nov 17, 2024 at 4:49 am



You asked:

Solve the equation \({z}^{5} = -32\) for the variable \(z\).

MathBot Answer:

The solutions are given by the formula: \[z = 2 \cos{\left(\frac{\pi}{5} + \frac{2 \pi k}{5} \right)} + 2 \sin{\left(\frac{\pi}{5} + \frac{2 \pi k}{5} \right)} i\] where \(k\) is an integer in \(0 \le k < 5\).

There is \(1\) real solution.

There are \(4\) solutions with nonzero imaginary part.

Real solutions: \[\begin{aligned}z &= -2\end{aligned}\]

Solutions with nonzero imaginary part: \[\begin{aligned}z &= \frac{1}{2} + \frac{\sqrt{5}}{2} + 2 i \sqrt{\frac{5}{8} - \frac{\sqrt{5}}{8}} \\&\approx 1.618034 + 1.1755705 i\\z &= - \frac{\sqrt{5}}{2} + \frac{1}{2} + 2 i \sqrt{\frac{\sqrt{5}}{8} + \frac{5}{8}} \\&\approx -0.61803399 + 1.902113 i\\z &= - \frac{\sqrt{5}}{2} + \frac{1}{2} - 2 i \sqrt{\frac{\sqrt{5}}{8} + \frac{5}{8}} \\&\approx -0.61803399 -1.902113 i\\z &= \frac{1}{2} + \frac{\sqrt{5}}{2} - 2 i \sqrt{\frac{5}{8} - \frac{\sqrt{5}}{8}} \\&\approx 1.618034 -1.1755705 i\end{aligned}\]

\(i\) is the imaginary unit, defined as \(i^2 = -1\).

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