z^6+1=0;

asked by guest
on Oct 24, 2024 at 12:17 pm



You asked:

Solve the equation \({z}^{6} + 1 = 0\) for the variable \(z\).

MathBot Answer:

The solutions are given by the formula: \[z = \cos{\left(\frac{\pi}{6} + \frac{\pi k}{3} \right)} + \sin{\left(\frac{\pi}{6} + \frac{\pi k}{3} \right)} i\] where \(k\) is an integer in \(0 \le k < 6\).

There are \(0\) real solutions.

There are \(6\) solutions with nonzero imaginary part.

The are no real solutions.

Solutions with nonzero imaginary part: \[\begin{aligned}z &= \frac{\sqrt{3}}{2} + \frac{i}{2} \\&\approx 0.8660254 + 0.5 i\\z &= i\\z &= - \frac{\sqrt{3}}{2} + \frac{i}{2} \\&\approx -0.8660254 + 0.5 i\\z &= - \frac{\sqrt{3}}{2} - \frac{i}{2} \\&\approx -0.8660254 -0.5 i\\z &= -i\\z &= \frac{\sqrt{3}}{2} - \frac{i}{2} \\&\approx 0.8660254 -0.5 i\end{aligned}\]

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

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