Question

How do you write the following expression in standard form$\dfrac{i }{3-2i}+\dfrac{2i }{3+8i }$?

Answer 1

Hint: To solve the above question we will use simple mathematical operations. The above given question is a complex number. A complex number is a number that can be expressed in the form of$a+i b$, where $a,b$ are real numbers and $i $ represents the imaginary part of the complex number. Real value is a number which has some real or discrete or possible value. But imaginary value is the number with a real number multiplied with an imaginary part $i $.

Complete step by step solution:
The given equation is:
 $\Rightarrow \dfrac{i }{3-2i }+\dfrac{2i }{3+8i }$
Now first we will take the LCM of the both the fraction in the given equation, then we get
$\begin{align}
  & \Rightarrow \dfrac{i \left( 3+8i \right)+2i \left( 3-2i \right)}{\left( 3-2i \right)\left( 3+8i \right)} \\
 & \Rightarrow \dfrac{3i +8{{i }^{2}}+6i -4{{i }^{2}}}{9+24i -6i -16{{i }^{2}}} \\
\end{align}$
Now we all know that the value of ${{i }^{2}}$ is equal to$-1$, by putting this we get
$\begin{align}
  & \Rightarrow \dfrac{9i -8+4}{9+18i +16} \\
 & \Rightarrow \dfrac{9i -4}{18i +15} \\
\end{align}$
Now we will simply multiply and divide the above equation with the conjugate of denominator of the above equation then we get
$\Rightarrow \dfrac{9i -4}{15+i 18}\times \dfrac{15-18i }{15-18i }$
Now we will use the algebraic property on the denominator part of the equation which is as$\Rightarrow \left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
Applying this on the above equation we get
$\begin{align}
  & \Rightarrow \dfrac{135i -162{{i }^{2}}-60+72i }{{{\left( 15 \right)}^{2}}-{{\left( 18i \right)}^{2}}} \\
 & \Rightarrow \dfrac{207i +162-60}{225+324} \\
 & \Rightarrow \dfrac{102+207i }{549} \\
\end{align}$
Now we will write the above equation in the split form, then we get
$\Rightarrow \dfrac{102}{549}+\dfrac{207}{549}i $
We can more simplify it,
$\Rightarrow \dfrac{34}{183}+\dfrac{23}{61}i $
Hence we get the simplified form of the given equation $\dfrac{i }{3-2i }+\dfrac{2i }{3+8i }$ is $\dfrac{34}{183}+\dfrac{23}{61}i $.

Note: We should know the standard form of writing an imaginary number, which is written as $a+i b$, where $a,b$ are the real numbers. We can go wrong to simplify these types of fractions by not multiplying the expression with the conjugate of the denominator.