Question

How do you simplify \[\dfrac{8i}{2-i}\] ?

Answer 1

Hint: This problem on rational numbers can be solved by rationalising the denominator; it is multiplying and dividing the given problem with the conjugate of the denominator. Now simplify the numerator and denominator using basic algebraic identities. After the simplification we get a complex number of the form \[a\]\[+\]\[ib\].

Complete step by step solution:
For simplification of the given problem,
\[\dfrac{8i}{2-i}\]
Firstly rationalise the denominator to remove the complex number from the denominator,
Now multiply and divide by conjugate of \[2-i\]
Since conjugate of \[a\]\[+\]\[ib\] is \[a\]\[-\]\[ib\]
Therefore Conjugate of \[2-i\] is \[2-(-i)\]\[=\] \[2+i\]
\[\dfrac{8i}{2-i}\]
\[\Rightarrow \] \[\dfrac{8i}{2-i}\]\[\times \]\[\dfrac{2+i}{2+i}\]
\[\Rightarrow \] \[\dfrac{8i(2+i)}{(2-i)(2+i)}\]
Using algebraic expression \[(a+b)(a-b)\]\[=\] \[({{a}^{2}}-{{b}^{2}})\]
\[\Rightarrow \] \[\dfrac{16i+8{{i}^{2}}}{({{2}^{2}}-{{i}^{2}})}\]
We know, \[{{i}^{2}}\] \[=\] \[i\times i\] \[=\] \[\sqrt{-1}\] \[\times \]\[\sqrt{-1}\]\[=\] \[{{\left( \sqrt{-1} \right)}^{2}}\]\[=\]\[\left( -1 \right)\] and \[{{2}^{2}}=2\times 2=4\], so we get
\[\Rightarrow \] \[\dfrac{16i+8(-1)}{(4-(-1))}\]
\[\Rightarrow \] \[\dfrac{16i-8}{(4+1)}\]
Simplifying further, we get
\[\Rightarrow \] \[\dfrac{-8+16i}{5}\]
\[\Rightarrow \] \[\dfrac{8(-1+2i)}{5}\]
\[\Rightarrow \] \[\dfrac{8}{5}(-1+2i)\]
We can write it also as,
\[\Rightarrow \] \[-\dfrac{8}{5}\] \[+\]\[\dfrac{16i}{5}\]
So, when we simplify the given problem \[\dfrac{8i}{2-i}\] , we obtain the simplified answer as \[\dfrac{8}{5}(-1+2i)\]

Note: While solving this problem we should have the knowledge of finding the conjugate of complex numbers because it’s very essential to simplify the fraction with a complex number in the denominator. It is required to have the knowledge of algebraic identities to simplify the given equation. The major point to be remembered while solving this type of question is to always multiply and divide by the conjugate of the denominator term and not the numerator, so this process is called rationalising the denominator. Always keep the final answer in the standard form of representation of the complex number it is as \[a+ib\] to avoid the confusion between the real part and imaginary part of the complex number.