Question

How do you simplify \[\dfrac{5-i}{2-i}-\dfrac{3-7i}{2-3i}\] and write in \[a+bi\] form?

Answer 1

Hint: In this problem, we have to simplify the above complex expression. We can first take conjugate for both the fractions and multiply both numerator and the denominator with the conjugate to get a \[a+bi\] form. We can find the conjugate by changing the sign of the imaginary part. We can then calculate and simplify the remaining terms to get the final answer.

Complete step by step solution:
We know that the given complex expression to be simplified is,
\[\Rightarrow \dfrac{5-i}{2-i}-\dfrac{3-7i}{2-3i}\]……. (1)
We can now take the complex conjugate for the denominator in the first fraction, we get
The complex conjugate of \[2-i\] is \[2+i\] .
We can now take the complex conjugate for the denominator in the second fraction, we get
The complex conjugate of \[2-3i\] is \[2+3i\] .
We can now multiply the both complex conjugates respectively in the expression (1), we get
\[\Rightarrow \dfrac{5-i}{2-i}\times \dfrac{2+i}{2+i}-\dfrac{3-7i}{2-3i}\times \dfrac{2+3i}{2+3i}\]
We can now simplify the above step by multiplying, we get
\[\begin{align}
  & \Rightarrow \dfrac{\left( 5-i \right)\left( 2+i \right)}{\left( 2-i \right)\left( 2+i \right)}-\dfrac{\left( 3-7i \right)\left( 2+3i \right)}{\left( 2-3i \right)\left( 2+3i \right)} \\
 & \Rightarrow \dfrac{10+3i-{{i}^{2}}}{{{2}^{2}}-{{i}^{2}}}-\dfrac{6-5i-21{{i}^{2}}}{{{2}^{2}}-{{\left( 3i \right)}^{2}}} \\
\end{align}\]
We know that \[{{i}^{2}}=-1\], using this we can simplify the above step, we get
\[\begin{align}
  & \Rightarrow \dfrac{10+3i+1}{5}-\dfrac{6-5i+21}{13} \\
 & \Rightarrow \dfrac{11+3i}{5}-\dfrac{27-5i}{13} \\
\end{align}\]
We can now further simplify the above step, we get
\[\begin{align}
  & \Rightarrow \dfrac{11}{5}-\dfrac{27}{13}+\left( \dfrac{3}{5}+\dfrac{5}{13} \right)i \\
 & \Rightarrow \dfrac{8}{65}+\dfrac{64}{65}i \\
\end{align}\]
Therefore, the simplified form is \[\dfrac{8}{65}+\dfrac{64}{65}i\].

Note: Students make mistakes while taking conjugate of the denominator, we should remember that the conjugate is changing the sign of the imaginary part. We should also remember that \[{{i}^{2}}=-1\]. We will make mistakes while writing positive and negative signs while substituting \[{{i}^{2}}=-1\], in which we should concentrate.