Answer
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Hint: Here in this question, we have to simplify the given question. The question is in the form of fraction, where both numerator and denominator contain imaginary numbers. When the number contains an imaginary number then the number is a complex number. Hence by using the arithmetic operations we obtain the required solution.
Complete step-by-step answer:
In mathematics we have different forms and kinds of numbers. The complex number is a combination of both the real part and the imaginary part. The complex number is represented as \[(a \pm ib)\]
Since the given number contains the imaginary part we can rationalise the number by multiplying its conjugate.
Now consider the given question \[\dfrac{{2i}}{{1 - i}}\]
We rationalise the denominator since both numerator and denominator contain imaginary numbers. So multiply and divide the number by (1+i) we have
\[ \Rightarrow \dfrac{{2i}}{{1 - i}} \times \dfrac{{1 + i}}{{1 + i}}\]
On multiplying we get
\[ \Rightarrow \dfrac{{2i(1 + i)}}{{(1 - i)(1 + i)}}\]
The term in the denominator is in the form of (a + b) (a-b), by applying the standard algebraic formula we have
\[ \Rightarrow \dfrac{{2i + 2{i^2}}}{{{1^2} - {i^2}}}\]
As we know that \[{i^2} = - 1\] , by applying this we have
\[ \Rightarrow \dfrac{{2i + 2( - 1)}}{{1 - ( - 1)}}\]
On simplifying we get
\[ \Rightarrow \dfrac{{2i - 2}}{{1 + 1}}\]
On further simplifying we get
\[ \Rightarrow \dfrac{{2i - 2}}{2}\]
Take 2 as a common in the numerator we have
\[ \Rightarrow \dfrac{{2(i - 1)}}{2}\]
On cancelling the 2 from both numerator and denominator we have
\[ \Rightarrow i - 1\]
Hence, we have simplified the given number and found the value.
So, the correct answer is “i - 1”.
Note: The rationalising the number is making the number as a finite. In most of the cases we rationalise the number for the number containing the square root. By rationalising the imaginary number, we get a finite number. Rationalising is applicable mostly to the fraction number or square root number.
Complete step-by-step answer:
In mathematics we have different forms and kinds of numbers. The complex number is a combination of both the real part and the imaginary part. The complex number is represented as \[(a \pm ib)\]
Since the given number contains the imaginary part we can rationalise the number by multiplying its conjugate.
Now consider the given question \[\dfrac{{2i}}{{1 - i}}\]
We rationalise the denominator since both numerator and denominator contain imaginary numbers. So multiply and divide the number by (1+i) we have
\[ \Rightarrow \dfrac{{2i}}{{1 - i}} \times \dfrac{{1 + i}}{{1 + i}}\]
On multiplying we get
\[ \Rightarrow \dfrac{{2i(1 + i)}}{{(1 - i)(1 + i)}}\]
The term in the denominator is in the form of (a + b) (a-b), by applying the standard algebraic formula we have
\[ \Rightarrow \dfrac{{2i + 2{i^2}}}{{{1^2} - {i^2}}}\]
As we know that \[{i^2} = - 1\] , by applying this we have
\[ \Rightarrow \dfrac{{2i + 2( - 1)}}{{1 - ( - 1)}}\]
On simplifying we get
\[ \Rightarrow \dfrac{{2i - 2}}{{1 + 1}}\]
On further simplifying we get
\[ \Rightarrow \dfrac{{2i - 2}}{2}\]
Take 2 as a common in the numerator we have
\[ \Rightarrow \dfrac{{2(i - 1)}}{2}\]
On cancelling the 2 from both numerator and denominator we have
\[ \Rightarrow i - 1\]
Hence, we have simplified the given number and found the value.
So, the correct answer is “i - 1”.
Note: The rationalising the number is making the number as a finite. In most of the cases we rationalise the number for the number containing the square root. By rationalising the imaginary number, we get a finite number. Rationalising is applicable mostly to the fraction number or square root number.
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