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Question

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

How do you simplify 8i2i\dfrac{8i}{2-i} ?

Explanation

Solution

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,
8i2i\dfrac{8i}{2-i}
Firstly rationalise the denominator to remove the complex number from the denominator,
Now multiply and divide by conjugate of 2i2-i
Since conjugate of a$$$$+$$$$ib is a$$$$-$$$$ib
Therefore Conjugate of 2i2-i is 2-(-i)$$$$= 2+i2+i
8i2i\dfrac{8i}{2-i}
\Rightarrow \dfrac{8i}{2-i}$$$$\times $$$$\dfrac{2+i}{2+i}
\Rightarrow 8i(2+i)(2i)(2+i)\dfrac{8i(2+i)}{(2-i)(2+i)}
Using algebraic expression (a+b)(a-b)$$$$= (a2b2)({{a}^{2}}-{{b}^{2}})
\Rightarrow 16i+8i2(22i2)\dfrac{16i+8{{i}^{2}}}{({{2}^{2}}-{{i}^{2}})}
We know, i2{{i}^{2}} == i×ii\times i == 1\sqrt{-1} \times $$$$\sqrt{-1}$$$$= {{\left( \sqrt{-1} \right)}^{2}}$$$$=$$$$\left( -1 \right) and 22=2×2=4{{2}^{2}}=2\times 2=4, so we get
\Rightarrow 16i+8(1)(4(1))\dfrac{16i+8(-1)}{(4-(-1))}
\Rightarrow 16i8(4+1)\dfrac{16i-8}{(4+1)}
Simplifying further, we get
\Rightarrow 8+16i5\dfrac{-8+16i}{5}
\Rightarrow 8(1+2i)5\dfrac{8(-1+2i)}{5}
\Rightarrow 85(1+2i)\dfrac{8}{5}(-1+2i)
We can write it also as,
\Rightarrow 85-\dfrac{8}{5} +$$$$\dfrac{16i}{5}
So, when we simplify the given problem 8i2i\dfrac{8i}{2-i} , we obtain the simplified answer as 85(1+2i)\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+iba+ib to avoid the confusion between the real part and imaginary part of the complex number.