Question

one of the solutions of equation z^3=w is z1=-5√3+5i what is one of the other two solutions if w is a complex number​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

The other two solution are $\bold{5\sqrt{3} +5i}$  and  $\bold{-10i}$.

Step-by-step explanation:

Given that $\bold{z^3=\omega}$ and $z_1=-5\sqrt{3} +5i$

Now the complex cube roots of unity $\omega$ satisfies  $\omega\neq 1,\omega^{3}=1.$

$\omega^3=1\\\omega^3-1=0\\(\omega-1)(\omega^2+\omega+1)=0$

The first factor is presumed non-zero so we have

$\omega=\frac{1}{2} (-1\pm i\sqrt{3})$

are the two complex cube roots of unity. Usually we assign

$\omega=\frac{1}{2} (-1\pm i\sqrt{3} )$

and the other is $\omega^x=\omega^2$

So if $z_1$ is one root of $z^3=a$ the other two are $z_1\omega$ and $z_1\omega^x$ because

$(z_1\omega)^3=z_{1}^{3}\omega^3-a(1)-a$

and similarly for $z_1\omega^x$ is

$z_1=5(-\sqrt{3} +i)$

$z_2=z_1\omega\\=5(-\sqrt{3} -i)\frac{1}{2}(-1+i\sqrt{3} )\\=\frac{5}{2} (\sqrt{3} -\sqrt{3} -i-3i)\\=-10i$

$z_3=z_1\omega^x\\=\frac{5}{2}(-\sqrt{3} +i)(-1-i\sqrt{3})(-1-i\sqrt{3} )\\ =\frac{5}{2}(\sqrt{3} +\sqrt{3} -i+3i)\\=5(\sqrt{3}+i)$

Therefore, The other two solution are $\bold{5\sqrt{3} +5i}$  and  $\bold{-10i}$.

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Q- if $z^3=w$ is $z_1=-5\sqrt{3}+5i$ then what is one of other two solution if w is a complex numbers.

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