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Question: Find the complex number \(\omega\) satisfying the equation \({{z}^{3}}=8i\) and lying in the second ...

Find the complex number ω\omega satisfying the equation z3=8i{{z}^{3}}=8i and lying in the second quadrant on the complex plane.
A. i+3i+\sqrt{3} lies in the first quadrant.
B. i−3i-\sqrt{3} lies in the second quadrant.
C. −i−3-i-\sqrt{3} lies in the third quadrant.
D. −i+3-i+\sqrt{3} lies in the fourth quadrant.

Explanation

Solution

Firstly, convert the given expression containing the complex number, z3=8i{{z}^{3}}=8i in such a way that we will be left with one on the RHS. We do this because omega ω\omega is easier to represent in terms of 1. Now evaluate the same further to get three terms out of which check which of the options lies in the second quadrant.

Complete step-by-step solution:
Firstly, let us rearrange the terms in the given expression containing the complex number, z3=8i{{z}^{3}}=8i
⇒z3=8i\Rightarrow {{z}^{3}}=8i
Now multiply and divide with i2{{i}^{2}} on the RHS.
⇒z3=8i×i2i2\Rightarrow {{z}^{3}}=8i\times \dfrac{{{i}^{2}}}{{{i}^{2}}}
Now evaluate the numerator on the RHS.
⇒z3=8i3i2\Rightarrow {{z}^{3}}=\dfrac{8{{i}^{3}}}{{{i}^{2}}}
Now,
⇒z3=8i3−1\Rightarrow {{z}^{3}}=\dfrac{8{{i}^{3}}}{-1}
Now divide the entire expression with 8i38{{i}^{3}}
⇒z38i3=−1\Rightarrow \dfrac{{{z}^{3}}}{8{{i}^{3}}}=-1
Now make the power common to the entire fraction.
And also negate the entire expression.
⇒(z−2i)3=1\Rightarrow {{\left( \dfrac{z}{-2i} \right)}^{3}}=1
Now cube root on both sides of the equation.
⇒z−2i=13\Rightarrow \dfrac{z}{-2i}=\sqrt[3]{1}
According to the complex numbers on a plane, we know that the roots of the expression,
13\sqrt[3]{1} are 1,ω,ω21,\omega ,{{\omega }^{2}}
Hence, we shall get three cases as below.
We can rewrite the above expression as,
⇒z−2i=1\Rightarrow \dfrac{z}{-2i}=1 or ω\omega or ω2{{\omega }^{2}}
Now bring the -2i to the RHS of the expression.
⇒z=−2i\Rightarrow z=-2i or −2iω-2i\omega or −2iω2-2i{{\omega }^{2}}
We also know the values of ω=−1+3i2\omega =\dfrac{-1+\sqrt{3}i}{2} and that of ω2=1−3i2{{\omega }^{2}}=\dfrac{1-\sqrt{3}i}{2}
Upon substituting these values, we get,
⇒z=−2i\Rightarrow z=-2i or −2i( −1+3i2)-2i\left( ~\dfrac{-1+\sqrt{3}i}{2} \right) or −2i(1−3i2)-2i\left( \dfrac{1-\sqrt{3}i}{2} \right)
Now evaluate further.
⇒z=−2i\Rightarrow z=-2i or 3+i\sqrt{3}+i or −3+i-\sqrt{3}+i
Now out of these three roots, −3+i-\sqrt{3}+i lies in the second quadrant according to the complex plane.
Hence, option B is correct.

Note: The complex number z=x+iyz=x+iy can be represented on the plane as the coordinates, (x,y)\left( x,y \right). Given that xx is the real part of the complex number and yy is the imaginary part of the complex number.