
How do you simplify \[\dfrac{{2 + i}}{{1 - i}}\] completely?
Answer
561.9k+ views
Hint: In order to convert the above question into standard we have to multiply both denominator and numerator with the complex conjugate of the denominator .Conjugate of the denominator will be \[1 + i\] and multiplying this with both numerator and denominator will give implication of the complex fraction .
\[{(A + B)^2} = {A^2} + {B^2} + 2 \times A \times B\]
$ (A + B)(A - B) = {A^2} - {B^2} $
$
{i^2} = - 1 \\
{i^3} = - i \;
$
Complete step-by-step answer:
Given a complex fraction \[\dfrac{{2 + i}}{{1 - i}}\] .let it be z
Here i is the imaginary number
First we’ll find out the complex conjugate of denominator of z i.e. \[1 - i\] which will be \[1 + i\]
Now dividing both numerator and denominator with the complex conjugate of denominator
\[ = \dfrac{{2 + i}}{{1 - i}} \times \dfrac{{1 + i}}{{1 + i}}\]
Using formula $ (A + B)(A - B) = {A^2} - {B^2} $
\[ = \dfrac{{\left( {2 + i} \right) \times (1 + i)}}{{{1^2} - {i^2}}}\]
Using formula \[{(A + B)^2} = {A^2} + {B^2} + 2 \times A \times B\]
\[ = \dfrac{{2 + 3i + {i^2}}}{{1 - {i^2}}}\]
Replacing $ {i^2}\, $ with $ - 1 $
\[
= \dfrac{{2 + 3i + ( - 1)}}{{1 - ( - 1)}} \\
= \dfrac{{2 + 3i - 1}}{{1 + 1}} \\
= \dfrac{{1 + 3i}}{2} \;
\]
Converting the above question into standard form by comparing it with standard form $ a + ib $
\[ = \dfrac{1}{2} + \dfrac{{3i}}{2}\]
Where Real number is \[\dfrac{1}{2}\] and imaginary number is \[\dfrac{{3i}}{2}\]
Therefore ,our required answer is \[ = \dfrac{1}{2} + \dfrac{{3i}}{2}\]
So, the correct answer is “\[ = \dfrac{1}{2} + \dfrac{{3i}}{2}\]”.
Note: 2. A Complex number is a number which are expressed in the form $ a + ib $ where $ ib $ is the imaginary part and $ a $ is the real number .i is generally known by the name iota, or in simple words complex numbers are the combination of a real number and an imaginary number .
3. Complex numbers are very useful in representing periodic motion like water waves, light waves and current and many more things which depend on sine or cosine waves.
4.The Addition or multiplication of any 2-conjugate complex number always gives an answer which is a real number.
\[{(A + B)^2} = {A^2} + {B^2} + 2 \times A \times B\]
$ (A + B)(A - B) = {A^2} - {B^2} $
$
{i^2} = - 1 \\
{i^3} = - i \;
$
Complete step-by-step answer:
Given a complex fraction \[\dfrac{{2 + i}}{{1 - i}}\] .let it be z
Here i is the imaginary number
First we’ll find out the complex conjugate of denominator of z i.e. \[1 - i\] which will be \[1 + i\]
Now dividing both numerator and denominator with the complex conjugate of denominator
\[ = \dfrac{{2 + i}}{{1 - i}} \times \dfrac{{1 + i}}{{1 + i}}\]
Using formula $ (A + B)(A - B) = {A^2} - {B^2} $
\[ = \dfrac{{\left( {2 + i} \right) \times (1 + i)}}{{{1^2} - {i^2}}}\]
Using formula \[{(A + B)^2} = {A^2} + {B^2} + 2 \times A \times B\]
\[ = \dfrac{{2 + 3i + {i^2}}}{{1 - {i^2}}}\]
Replacing $ {i^2}\, $ with $ - 1 $
\[
= \dfrac{{2 + 3i + ( - 1)}}{{1 - ( - 1)}} \\
= \dfrac{{2 + 3i - 1}}{{1 + 1}} \\
= \dfrac{{1 + 3i}}{2} \;
\]
Converting the above question into standard form by comparing it with standard form $ a + ib $
\[ = \dfrac{1}{2} + \dfrac{{3i}}{2}\]
Where Real number is \[\dfrac{1}{2}\] and imaginary number is \[\dfrac{{3i}}{2}\]
Therefore ,our required answer is \[ = \dfrac{1}{2} + \dfrac{{3i}}{2}\]
So, the correct answer is “\[ = \dfrac{1}{2} + \dfrac{{3i}}{2}\]”.
Note: 2. A Complex number is a number which are expressed in the form $ a + ib $ where $ ib $ is the imaginary part and $ a $ is the real number .i is generally known by the name iota, or in simple words complex numbers are the combination of a real number and an imaginary number .
3. Complex numbers are very useful in representing periodic motion like water waves, light waves and current and many more things which depend on sine or cosine waves.
4.The Addition or multiplication of any 2-conjugate complex number always gives an answer which is a real number.
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