Question

How do you factor ${{a}^{2}}+{{b}^{2}}$?

Answer 1

Hint:
In this problem we have to find factors of the ${{a}^{2}}+{{b}^{2}}$. We cannot find factors given expression. So, we will use the concept of the complex numbers. We will expand the given expression as ${{a}^{2}}-\left( -{{b}^{2}} \right)$. Now we will use the well known formula in complex numbers which is $i=\sqrt{-1}$. Now we will simplify the obtained equation by using the algebraic formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. After applying this formula, we will use some substitutions according to the concept of complex numbers which are $z=a+ib$, $\overline{z}=a-bi$ to get the factors.

FORMULA USE:
1. ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ .
2. $i=\sqrt{-1}$
3. $z=a+ib$, $\overline{z}=a-bi$

Complete step by step solution:
Given that, ${{a}^{2}}+{{b}^{2}}$ .
We can’t find the factor of above with the given expression. We will use the concept of complex numbers
Now we will expand given expression by taking a negative sign common from the term $+{{b}^{2}}$, then we will get
${{a}^{2}}+{{b}^{2}}={{a}^{2}}-\left( -{{b}^{2}} \right)$
In complex numbers we have the basic formula
$i=\sqrt{-1}$.
Squaring on both sides of the above equation, then we will get
${{i}^{2}}=-1$
Substituting this value in the equation ${{a}^{2}}+{{b}^{2}}={{a}^{2}}-\left( -{{b}^{2}} \right)$, then we will get
$\Rightarrow {{a}^{2}}+{{b}^{2}}={{a}^{2}}-{{\left( bi \right)}^{2}}$ .
We have the algebraic formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ . Applying this formula in the above equation, then we will get
${{a}^{2}}+{{b}^{2}}=\left( a+ib \right)\left( a-ib \right)$ .
If we assume the number $\left( a+ib \right)$ as $z$, then $a-ib$ should be $\bar{z}$.
So, we can write the above equation by assuming $z=a+ib$, then we will get
${{a}^{2}}+{{b}^{2}}=z\times \overline{z}$
Now we will simplify above expression then
$\begin{align}
  & \Rightarrow {{a}^{2}}+{{b}^{2}}={{z}^{2}} \\
 & \\
\end{align}$
Where $z$ is $z=a+ib$


Note:
The value of ${{a}^{2}}+{{b}^{2}}$ also calculated from the algebraic formulas ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$. From both the formulas we can write the value of ${{a}^{2}}+{{b}^{2}}$ as
${{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab$,
${{a}^{2}}+{{b}^{2}}={{\left( a-b \right)}^{2}}+2ab$,
${{a}^{2}}+{{b}^{2}}=\dfrac{{{\left( a+b \right)}^{2}}+{{\left( a-b \right)}^{2}}}{2}$.