If $a, b \in R$ and $i=\sqrt{-1}$, then the number of ordered pairs of real numbers $(a, b)$ satisfying the…
If $a, b \in R$ and $i=\sqrt{-1}$, then the number of ordered pairs of real numbers $(a, b)$ satisfying the condition $(a+b i)^3=a-b i$ is
- 3
- 2
- 4
- 5
Solution
Given condition,
$
\begin{aligned}
& (a+b i)^3=a-b i \\
& \text { i.e. } a^3+i^3 b^3+3 a^2 b i-3 a b^2 i^2=a-b i \\
& \Rightarrow \quad a^3-i b^3+i 3 a^2 b-3 a b^2=a-b i \\
& \left(a^3-3 a b^2\right)-i\left(b^3-3 a^2 b\right)=a-b i \\
&
\end{aligned}
$
Equating the coefficient, we obtain
$
\begin{aligned}
a^3-3 a b^2 & =a \text { and } b^3-3 a^2 b=b \\
\Rightarrow a\left(a^2-3 b^2\right) & =a \text { and } b\left(b^2-3 a^2\right)=b \\
\Rightarrow \quad & a^2-3 b^2=1 \\
\text { and } b^2-3 a^2 & =1
\end{aligned}
$
Use Eq. (i) in (ii)
$
\begin{aligned}
b^2-3\left(1+3 b^2\right) & =1 \\
\Rightarrow \quad b^2-3-9 b^2 & =1 \text { or } 8 b^2+4=0
\end{aligned}
$
This gives $b^2=-1 / 2$
$
\therefore \quad b= \pm \frac{i}{\sqrt{2}}
$
Now,
$
\begin{aligned}
a^2 & =1+3 b^2=1+3\left(\frac{-1}{2}\right)=1-\frac{3}{2}=-\frac{1}{2} \\
\therefore \quad a & = \pm \frac{i}{\sqrt{2}}
\end{aligned}
$
$\therefore$ Possible ordered pairs are
$
\begin{aligned}
& \left(\frac{i}{\sqrt{2}}, \frac{i}{\sqrt{2}}\right),\left(-\frac{i}{\sqrt{2}},-\frac{i}{\sqrt{2}}\right),\left(\frac{i}{\sqrt{2}}, \frac{-i}{\sqrt{2}}\right), \\
& \left(-\frac{i}{\sqrt{2}}, \frac{i}{\sqrt{2}}\right)
\end{aligned}
$
Number of ordered pair $=4$
Asked in: AP EAMCET 2021 (23 Aug Shift 1)
Practice more Complex Number questions on Aicharya