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Solve for z: \(\large i{ z }^{ 3 }+z^{ 2 }-z+i=0\)

 Jun 29, 2020
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\(z^2 (iz + 1) + i(iz + 1) = 0\\ (z^2 + i)(iz + 1) = 0\\ z^2 = -i \text{ or } z = \dfrac{-1}{i} = i\\ \)

 

For z2 = -i, let \(z = re^{i\theta}\).

 

\(r^2 e^{2i\theta} = 1\cdot e^{3i\pi/2}\\ r = 1\\ \theta = \dfrac{3\pi}4 \text{ or } \theta = \dfrac{7\pi}4\)

 

Therefore the other two solutions are \(z = e^{3i\pi/4}\) and \(z = e^{7i\pi/4}\), which is \(-\dfrac{\sqrt 2}2 + \dfrac{\sqrt 2}{2} i\) and \(\dfrac{\sqrt 2}2 - \dfrac{\sqrt 2}{2} i\).

 Jun 29, 2020

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