Question

How do you simplify $\dfrac{2i}{3-4i}$?

Answer 1

Now to simplify the expression we will first rationalize the denominator by multiplying the numerator and denominator with $3+4i$ . Now using the property ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ . Now on simplifying the expression we will write the expression in a general form of complex number which is $a+ib$ where a and b are real numbers and $i=\sqrt{-1}$.

Complete step by step solution:
Now consider the given expression $\dfrac{2i}{3-4i}$ .
Now we know the given number is a complex number where $i=\sqrt{-1}$ .
Now we know that in real numbers the root of negative numbers are not defined. Hence we define complex numbers where $\sqrt{-1}$ is written as i. The general form of a complex number is $a+ib$ where a and b are both real numbers. Now we will try to write the given expression in general form of complex numbers.
Now to simplify the given expression we will first rationalize the denominator. Hence we will multiply the numerator and denominator by $3+4i$ . Hence we get,
$\Rightarrow \dfrac{2i\left( 3+4i \right)}{\left( 3+4i \right)\left( 3-4i \right)}$
Now we know that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ . Hence using this formula we get,
$\Rightarrow \dfrac{6i+8{{\left( i \right)}^{2}}}{{{3}^{2}}-{{\left( 4i \right)}^{2}}}$
Now we know that ${{i}^{2}}=-1$ Hence we get,
$\begin{aligned} & \\\ & \Rightarrow \dfrac{6i-8}{9+4} \\\ \end{aligned}$
$\begin{aligned} & \Rightarrow \dfrac{6i-8}{13} \\\ & \Rightarrow \dfrac{-8}{13}+\dfrac{6}{13}i \\\ \end{aligned}$
Now the complex number is in its general form which is $a+ib$ where a and b are real numbers. Hence the given expression is simplified.

Note: Now note that whenever we have a complex number in a denominator we first rationalize the fraction by multiplying the numerator and denominator by its complex conjugate. Hence we further simplify the expression and try to get the real number in the denominator.