If z in mathbb{C} and i z^3+4 z^2-z+4 i=0,the | vedclass

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(B) Given the complex equation: $i z^3+4 z^2-z+4 i=0$ Grouping the terms: $z^2(i z+4) - 1(z - 4i) = 0$ Wait,let us factorize correctly: $i z^3 + 4 z^2

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$\frac{1-i}{\sqrt{2}}$

Vedclass · Answer

(B) Given the complex equation: $i z^3+4 z^2-z+4 i=0$ Grouping the terms: $z^2(i z+4) - 1(z - 4i) = 0$ Wait,let us factorize correctly: $i z^3 + 4 z^2 - z + 4 i = 0$ $z^2(i z + 4) + i(i z + 4) = 0$ $(i z + 4)(z^2 + i) = 0$ This gives $z = \frac{-4}{i} = 4i$ or $z^2 = -i$. For $z^2 = -i$,we write $-i$ in polar form: $z^2 = \cos(\frac{3\pi}{2}) + i \sin(\frac{3\pi}{2})$. Using De Moivre's theorem,$z = \cos(\frac{3\pi}{4} + k\pi) + i \sin(\frac{3\pi}{4} + k\pi)$ for $k=0, 1$. For $k=0$,$z = \cos(\frac{3\pi}{4}) + i \sin(\frac{3\pi}{4}) = -\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}$. For $k=1$,$z = \cos(\frac{7\pi}{4}) + i \sin(\frac{7\pi}{4}) = \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}$. The magnitudes are $|4i| = 4$ and $|z| = \sqrt{(\pm \frac{1}{\sqrt{2}})^2 + (\mp \frac{1}{\sqrt{2}})^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = 1$. Thus,the roots with minimum magnitude are $\pm(\frac{1-i}{\sqrt{2}})$. Comparing with options,$\frac{1-i}{\sqrt{2}}$ is the correct choice.

If $z \in \mathbb{C}$ and $i z^3+4 z^2-z+4 i=0$,then a complex root of this equation having minimum magnitude is

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