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جستجو

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جستجوهای پرتکرار

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a) Find the roots of the equation

${z^2} + \left( {2\sqrt 3 } \right)z + 4 = 0$,

giving your answers in the form $x + iy$, where $x$ and $y$ are real.

b) State the modulus and argument of each root.

c) Showing all your working, verify that each root also satisfies the equation

${z^6} =  - 64$

پاسخ تشریحی :
نمایش پاسخ

a) Use the quadratic formula, completing the square, or the substitution $z = x + iy$ to find a root and use ${i^2} =  - 1$

Obtain final answers $ - \sqrt 3  \pm i$, or equivalent

b) State that the modulus of both roots is 2

State that the argument of $ - \sqrt 3  + i$ is ${150^ \circ }$ or $\frac{5}{6}\pi \left( {2.62} \right)$ radians

State that the argument of $ - \sqrt 3  - i$ is $ - {150^ \circ }$ (or ${210^ \circ }$) or $ - \frac{5}{6}\pi \left( { - 2.62} \right)$ radians or $\frac{7}{6}\pi \left( {3.67} \right)$ radians

c) Carry out an attempt to find the sixth power of a root

Verify that one of the roots satisfies ${z^6} =  - 64$

Verify that the other root satisfies the equation

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