
Find the complex number $\omega$ satisfying the equation ${{z}^{3}}=8i$ and lying in the second quadrant on the complex plane.
A. $i+\sqrt{3}$ lies in the first quadrant.
B. $i-\sqrt{3}$ lies in the second quadrant.
C. $-i-\sqrt{3}$ lies in the third quadrant.
D. $-i+\sqrt{3}$ lies in the fourth quadrant.
Answer
543.3k+ views
Hint: Firstly, convert the given expression containing the complex number, ${{z}^{3}}=8i$ in such a way that we will be left with one on the RHS. We do this because omega $\omega$ 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, ${{z}^{3}}=8i$
$\Rightarrow {{z}^{3}}=8i$
Now multiply and divide with ${{i}^{2}}$ on the RHS.
$\Rightarrow {{z}^{3}}=8i\times \dfrac{{{i}^{2}}}{{{i}^{2}}}$
Now evaluate the numerator on the RHS.
$\Rightarrow {{z}^{3}}=\dfrac{8{{i}^{3}}}{{{i}^{2}}}$
Now,
$\Rightarrow {{z}^{3}}=\dfrac{8{{i}^{3}}}{-1}$
Now divide the entire expression with $8{{i}^{3}}$
$\Rightarrow \dfrac{{{z}^{3}}}{8{{i}^{3}}}=-1$
Now make the power common to the entire fraction.
And also negate the entire expression.
$\Rightarrow {{\left( \dfrac{z}{-2i} \right)}^{3}}=1$
Now cube root on both sides of the equation.
$\Rightarrow \dfrac{z}{-2i}=\sqrt[3]{1}$
According to the complex numbers on a plane, we know that the roots of the expression,
$\sqrt[3]{1}$ are $1,\omega ,{{\omega }^{2}}$
Hence, we shall get three cases as below.
We can rewrite the above expression as,
$\Rightarrow \dfrac{z}{-2i}=1$ or $\omega$ or ${{\omega }^{2}}$
Now bring the -2i to the RHS of the expression.
$\Rightarrow z=-2i$ or $-2i\omega$ or $-2i{{\omega }^{2}}$
We also know the values of $\omega =\dfrac{-1+\sqrt{3}i}{2}$ and that of ${{\omega }^{2}}=\dfrac{1-\sqrt{3}i}{2}$
Upon substituting these values, we get,
$\Rightarrow z=-2i$ or $-2i\left( ~\dfrac{-1+\sqrt{3}i}{2} \right)$ or $-2i\left( \dfrac{1-\sqrt{3}i}{2} \right)$
Now evaluate further.
$\Rightarrow z=-2i$ or $\sqrt{3}+i$ or $-\sqrt{3}+i$
Now out of these three roots, $-\sqrt{3}+i$ lies in the second quadrant according to the complex plane.
Hence, option B is correct.
Note: The complex number $z=x+iy$ can be represented on the plane as the coordinates, $\left( x,y \right)$. Given that $x$ is the real part of the complex number and $y$ is the imaginary part of the complex number.
Complete step-by-step solution:
Firstly, let us rearrange the terms in the given expression containing the complex number, ${{z}^{3}}=8i$
$\Rightarrow {{z}^{3}}=8i$
Now multiply and divide with ${{i}^{2}}$ on the RHS.
$\Rightarrow {{z}^{3}}=8i\times \dfrac{{{i}^{2}}}{{{i}^{2}}}$
Now evaluate the numerator on the RHS.
$\Rightarrow {{z}^{3}}=\dfrac{8{{i}^{3}}}{{{i}^{2}}}$
Now,
$\Rightarrow {{z}^{3}}=\dfrac{8{{i}^{3}}}{-1}$
Now divide the entire expression with $8{{i}^{3}}$
$\Rightarrow \dfrac{{{z}^{3}}}{8{{i}^{3}}}=-1$
Now make the power common to the entire fraction.
And also negate the entire expression.
$\Rightarrow {{\left( \dfrac{z}{-2i} \right)}^{3}}=1$
Now cube root on both sides of the equation.
$\Rightarrow \dfrac{z}{-2i}=\sqrt[3]{1}$
According to the complex numbers on a plane, we know that the roots of the expression,
$\sqrt[3]{1}$ are $1,\omega ,{{\omega }^{2}}$
Hence, we shall get three cases as below.
We can rewrite the above expression as,
$\Rightarrow \dfrac{z}{-2i}=1$ or $\omega$ or ${{\omega }^{2}}$
Now bring the -2i to the RHS of the expression.
$\Rightarrow z=-2i$ or $-2i\omega$ or $-2i{{\omega }^{2}}$
We also know the values of $\omega =\dfrac{-1+\sqrt{3}i}{2}$ and that of ${{\omega }^{2}}=\dfrac{1-\sqrt{3}i}{2}$
Upon substituting these values, we get,
$\Rightarrow z=-2i$ or $-2i\left( ~\dfrac{-1+\sqrt{3}i}{2} \right)$ or $-2i\left( \dfrac{1-\sqrt{3}i}{2} \right)$
Now evaluate further.
$\Rightarrow z=-2i$ or $\sqrt{3}+i$ or $-\sqrt{3}+i$
Now out of these three roots, $-\sqrt{3}+i$ lies in the second quadrant according to the complex plane.
Hence, option B is correct.
Note: The complex number $z=x+iy$ can be represented on the plane as the coordinates, $\left( x,y \right)$. Given that $x$ is the real part of the complex number and $y$ is the imaginary part of the complex number.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

