Question

Find the real values of x and y for which
$\left( {{x}^{4}}+2xi \right)-\left( 3{{x}^{2}}+iy \right)=\left( 3-5i \right)+\left( 1+2iy \right)$

Answer 1

Hint: First of all transpose all the terms to LHS and separate the real and imaginary parts of the equation. Now equate the real and imaginary part of the equation to zero individually and get the real values of x and y by solving these equations.

Complete step-by-step solution -

In this question, we have to solve the equation $\left( {{x}^{4}}+2xi \right)-\left( 3{{x}^{2}}+iy \right)=\left( 3-5i \right)+\left( 1+2iy \right)$ for real values of x and y. First of all, let us consider the equation given in the question.
$\left( {{x}^{4}}+2xi \right)-\left( 3{{x}^{2}}+iy \right)=\left( 3-5i \right)+\left( 1+2iy \right)$
Let us transpose all the terms of the above equation to LHS, we get,
$\left( {{x}^{4}}+2xi \right)-\left( 3{{x}^{2}}+iy \right)-\left( 3-5i \right)-\left( 1+2iy \right)=0$
By separating real and imaginary terms of the above equation, we get,
$\left( {{x}^{4}}-3{{x}^{2}}-3-1 \right)+\left( 2x-y+5-2y \right)i=0$
$\Rightarrow \left( {{x}^{4}}-3{{x}^{2}}-4 \right)+\left( 2x+5-3y \right)i=0$
We know that in any equation the left-hand side is equal to the right-hand side. We can see that in the RHS of the above equation, both the real part and the imaginary part are zero. So, we can write the above equation as,
$\left( {{x}^{4}}-3{{x}^{2}}-4 \right)+\left( 2x+5-3y \right)i=0+0i.....\left( i \right)$
Now, by equating real part of the LHS to real part of the RHS of the above equation, we get,
${{x}^{4}}-3{{x}^{2}}-4=0$
We can write the above equation as
${{x}^{4}}-4{{x}^{2}}+{{x}^{2}}-4=0$
${{x}^{2}}\left( {{x}^{2}}-4 \right)+1\left( {{x}^{2}}-4 \right)=0$
$\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}-4 \right)=0$
So, we get ${{x}^{2}}=-1$ and ${{x}^{2}}=4$
Hence, we get $x=\pm i$ and $x=\pm 2$
We are given that x is real. So, we take only $x=\pm 2$.
Now by equating the imaginary part of the LHS to the imaginary part of the RHS of equation (i), we get,
$2x+5-3y=0.....\left( ii \right)$
By substituting x = 2 in the above equation, we get,
$2\left( 2 \right)+5-3y=0$
$\Rightarrow 4+5-3y=0$
$\Rightarrow 9-3y=0$
$\Rightarrow 3y=9$
$\Rightarrow y=\dfrac{9}{3}=3$
So, we get, y = 3 for x = 2.
Now, by substituting $x=-2$ in equation (ii), we get,
$2\left( -2 \right)+5-3y=0$
$\Rightarrow -4+5-3y=0$
$\Rightarrow 1-3y=0$
$\Rightarrow 3y=1$
$y=\dfrac{1}{3}$
So, we get, $y=\dfrac{1}{3}$ for x = – 2.
Hence, we get the values of (x, y) as (2, 3) and $\left( -2,\dfrac{1}{3} \right)$.

Note: In this question, students can cross check their answer by substituting x and y in the given equation as follows:
$\left( {{x}^{4}}+2xi \right)-\left( 3{{x}^{2}}+iy \right)=\left( 3-5i \right)+\left( 1+2iy \right)$
By substituting x = 2 and y = 3, we get,
$\left[ {{\left( 2 \right)}^{4}}+2.\left( 2 \right)i \right]-\left[ 3{{\left( 2 \right)}^{2}}+i\left( 3 \right) \right]=\left( 3-5i \right)+\left( 1+2-3i \right)$
$\Rightarrow \left( 16+4i \right)-\left( 12+3i \right)=\left( 3-5i \right)+\left( 1+6i \right)$
$\Rightarrow 4+i=4+i$
LHS = RHS
Here we get LHS = RHS. So, our answer is correct. Similarly, we can check for x = – 2 and $y=\dfrac{1}{3}$. Also, in this question, some students make this mistake of taking $x=\pm i$ which is wrong because we are given the question that values of x and y are real by $x=\pm i$ is imaginary. So, we only take $x=\pm 2$.