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Question: How do you divide \(\dfrac{{2i}}{{1 - i}}\)?...

How do you divide 2i1−i\dfrac{{2i}}{{1 - i}}?

Explanation

Solution

This problem deals with simplifying the complex numbers. A complex number is a number that can be expressed in the form of a+iba + ib, where aa and bb are real numbers, and ii represents the imaginary unit, satisfying the equation i2=−1{i^2} = - 1. Because no real number satisfies this equation, ii is called an imaginary number.

Complete step-by-step answer:
The given expression is 2i1−i\dfrac{{2i}}{{1 - i}}, we have to simplify the expression.
Now consider the given expression, as shown below:
⇒2i1−i\Rightarrow \dfrac{{2i}}{{1 - i}}
Now multiply and divide the above expression with 1+i1 + i, that is multiplying with this expression in the numerator and the denominator, as shown below:
⇒2i1−i×1+i1+i\Rightarrow \dfrac{{2i}}{{1 - i}} \times \dfrac{{1 + i}}{{1 + i}}
Now simplifying this expression by solving the expressions in the numerator and the denominator.
Consider the numerator as shown:
⇒2i(1+i)\Rightarrow 2i\left( {1 + i} \right)
Multiplying each term, as shown:
⇒2i(1+i)=2i+2i2\Rightarrow 2i\left( {1 + i} \right) = 2i + 2{i^2}
We know that i2=−1{i^2} = - 1, substituting this in the above expression as shown:
⇒2i(1+i)=2i−2\Rightarrow 2i\left( {1 + i} \right) = 2i - 2
Now taking the number 2 common on the right hand side of the equation, as shown:
∴2i(1+i)=2(i−1)\therefore 2i\left( {1 + i} \right) = 2\left( {i - 1} \right)
Now consider the denominator as shown:
⇒(1−i)(1+i)\Rightarrow \left( {1 - i} \right)\left( {1 + i} \right)
Multiplying each term, as shown:
⇒(1−i)(1+i)=1(1+i)−i(1+i)\Rightarrow \left( {1 - i} \right)\left( {1 + i} \right) = 1\left( {1 + i} \right) - i\left( {1 + i} \right)
⇒(1−i)(1+i)=1+i−i−i2\Rightarrow \left( {1 - i} \right)\left( {1 + i} \right) = 1 + i - i - {i^2}
Here ii and −i - i gets cancelled as shown below:
⇒(1−i)(1+i)=1−i2\Rightarrow \left( {1 - i} \right)\left( {1 + i} \right) = 1 - {i^2}
We know that i2=−1{i^2} = - 1, substituting this in the above expression as shown:
⇒(1−i)(1+i)=1−(−1)\Rightarrow \left( {1 - i} \right)\left( {1 + i} \right) = 1 - \left( { - 1} \right)
∴(1−i)(1+i)=2\therefore \left( {1 - i} \right)\left( {1 + i} \right) = 2
Now replacing the numerator and the denominator with the obtained expressions, as shown:
⇒2i1−i×1+i1+i=2(i−1)2\Rightarrow \dfrac{{2i}}{{1 - i}} \times \dfrac{{1 + i}}{{1 + i}} = \dfrac{{2\left( {i - 1} \right)}}{2}
Now 2 gets cancelled in both the numerator and the denominator, as shown:
⇒2i1−i×1+i1+i=i−1\Rightarrow \dfrac{{2i}}{{1 - i}} \times \dfrac{{1 + i}}{{1 + i}} = i - 1
The simplification of the expression 2i1−i\dfrac{{2i}}{{1 - i}} is given by:
∴2i1−i=i−1\therefore \dfrac{{2i}}{{1 - i}} = i - 1

Note:
Please note that the backbone of this new number system is the number ii, also known as the imaginary unit. So from here we can conclude that any imaginary number is also a complex number, and any real number is also a complex number.