Question

Let ${z_1}$ and ${z_2}$ be two roots of the equation ${z^2} + az + b = 0$, being complex. Further, assumed that the origin ${z_1}$ and ${z_2}$ from an equilateral triangle. Then, we find the equal value for the given equation.
a. ${a^2} = b$
b. ${a^2} = 2b$
c. ${a^2} = 3b$
d. ${a^2} = 4b$

Answer 1

Hint: Here we have to find the equal value of the given equation. First, we use the Equilateral triangle formula and Euler’s formula for complex numbers because the given equation is complex. Then, we simplify the equations and finally post the correct option from the given.

Formula used:
An equilateral triangle formula is
Euler’s formula is ${e^{i\pi }} = \cos \pi + i\sin \pi $

Complete answer:
Let ${z_1}$ and be two roots of the equation ${z^2} + az + b = 0$, being complex.
Assume that the origin ${z_1}$and ${z_2}$from an equilateral triangle.
An equilateral triangle formula is $x{e^{i\dfrac{\pi }{3}}} = y$
Here, $x$ and $y$ are complex numbers.
First we have, ${z_1} + {z_2} = - a$ and ${z_1}{z_2} = b$ from the given equation is ${z^2} + az + b = 0$
We use the equilateral triangle formula and we get,
${z_1}{e^{i\dfrac{\pi }{3}}} = {z_2}$
Here, ${z_1}$ and ${z_2}$ are complex numbers.
Then, we use the Euler’s formula is ${e^{i\pi }} = \cos \pi + i\sin \pi $ and we get,
${z_1}\left( {\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right) = {z_2}$
Here, $\pi $ value is $\dfrac{\pi }{3}$ because we use the equilateral triangle.
We use trigonometry table for $\cos \dfrac{\pi }{3} = \dfrac{1}{2}$and $\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$ and we get,
${z_2} = {z_1}\left( {\dfrac{1}{2} + \dfrac{{i\sqrt 3 }}{2}} \right)$
We take the LCM for the equation,
We get,
\[{z_2} = {z_1}\left( {\dfrac{{1 + i\sqrt 3 }}{2}} \right)\]
The cross multiplied the Right hand side by 2,
\[2{z_2} = {z_1}\left( {1 + i\sqrt 3 } \right)\]
Hence,
$2{z_2} = {z_1} + {z_1}i\sqrt 3 $
Now, keeping the variables on one side and complex number on other side. So we have to move ${z_1}i\sqrt 3 $ to the left hand side (LHS) and $2{z_2}$ to the right hand side (RHS). We get,
$2{z_2} - {z_1} = {z_1}i\sqrt 3 $
Squaring the above equation on both sides and simplifying that equation.
We get,
${\left( {2{z_2} - {z_1}} \right)^2} = {i^2}3{z_1}^2$
Now, we use ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ and ${{\text{i}}^2} = 1$ in the above equation. We get,
${2^2}{z_2}^2 + {z_1}^2 - 2\left( {2{z_1}{z_2}} \right) = - 3{z_1}^2$
$4{z_2}^2 + {z_1}^2 - 4{z_1}{z_2} = - 3{z_1}^2$
Separate the square terms in the above equation and we get,
$4{z_2}^2 + 3{z_1}^2 + {z_1}^2 = 4{z_1}{z_2}$
Now, adding the ${z_1}^2$ terms and we get,
$4{z_2}^2 + 4{z_1}^2 = 4{z_1}{z_2}$
$4\left( {{z_2}^2 + {z_1}^2} \right) = 4{z_1}{z_2}$
Hence,
 ${z_1}^2 + {z_2}^2 = {z_1}{z_2}$
We adding and subtraction by $2{z_1}{z_2}$ in the left hand side of the above equation and we get,
${z_1}^2 + {z_2}^2 + 2{z_1}{z_2} - 2{z_1}{z_2} = {z_1}{z_2}$
${\left( {{z_1} + {z_2}} \right)^2} - 2{z_1}{z_2} = {z_1}{z_2}$
Put ${z_1} + {z_2} = - a$ and ${z_1}{z_2} = b$ in the above equation and we get,
${\left( a \right)^2} - 2\left( b \right) = b$
${a^2} - 2b = b$
Rearranging the a and b terms and we get,
${a^2} = b + 2b$
Adding them, hence we get,
${a^2} = 3b$

Hence, the correct answer is option (C).

Note: We have to mine that, a complex number is a number that can be expressed in the form $a + bi$, where $a$ and $b$ are real number, and $i$ represents the imaginary unit, satisfying the equation ${i^2} = - 1$. Because no real number satisfies this equation, $i$ is called an imaginary number. The real number $a$ is called the real part of the complex number $a + bi$, the real number $b$ is called its imaginary part. To emphasize, the imaginary part does not include a factor $i$, that is the imaginary part is $b$ not $bi$. The set of complex numbers is denoted by either of the symbols $\mathbb{C}$ or $C$.