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Question: The imaginary part of \(\dfrac{{{\left( 1+i \right)}^{2}}}{i\left( 2i-1 \right)}\) \(\begin{aligne...

The imaginary part of (1+i)2i(2i−1)\dfrac{{{\left( 1+i \right)}^{2}}}{i\left( 2i-1 \right)}
(A)45 (B)0 (C)25 (D)−45 \begin{aligned} & \left( A \right)\dfrac{4}{5} \\\ & \left( B \right)0 \\\ & \left( C \right)\dfrac{2}{5} \\\ & \left( D \right)-\dfrac{4}{5} \\\ \end{aligned}

Explanation

Solution

From the above question we have to find the imaginary part of the complex number (1+i)2i(2i−1)\dfrac{{{\left( 1+i \right)}^{2}}}{i\left( 2i-1 \right)}. As we know that if a complex number is a+iba+ib then a is said to be the real part and b is said to be an imaginary part. First, we will multiply the terms in the denominator and then we will multiply both the numerator and denominator with the conjugate of denominator that is conjugate of a+iba+ib is a−iba-ib. and then we will expand the numerator term and we will multiply the numerator terms. Then we will get the imaginary part of the complex number.

Complete step by step solution:
From the question given that we have to find the imaginary part of
⇒(1+i)2i(2i−1)\Rightarrow \dfrac{{{\left( 1+i \right)}^{2}}}{i\left( 2i-1 \right)}
Now we will multiply the both terms in the numerator,
By multiplying we will get,
⇒(1+i)2i(2i−1)=(1+i)2(2(i)2−1(i))\Rightarrow \dfrac{{{\left( 1+i \right)}^{2}}}{i\left( 2i-1 \right)}=\dfrac{{{\left( 1+i \right)}^{2}}}{\left( 2{{\left( i \right)}^{2}}-1\left( i \right) \right)}
By further simplifying we will get,
⇒(1+i)2(−2−i)\Rightarrow \dfrac{{{\left( 1+i \right)}^{2}}}{\left( -2-i \right)}
Now we will multiply both the numerator and denominator with the conjugate of the denominator that is the conjugate of denominator (−2−i)\left( -2-i \right) is (−2+i)\left( -2+i \right)
By multiplying we will get,
⇒(1+i)2(−2−i)×(−2+i)(−2+i)\Rightarrow \dfrac{{{\left( 1+i \right)}^{2}}}{\left( -2-i \right)}\times \dfrac{\left( -2+i \right)}{\left( -2+i \right)}
⇒(1+i)2(−2+i)((−2)2−(i)2)\Rightarrow \dfrac{{{\left( 1+i \right)}^{2}}\left( -2+i \right)}{\left( {{\left( -2 \right)}^{2}}-{{\left( i \right)}^{2}} \right)}
By further simplifying we will get,
⇒(1+i)2(−2+i)(4−(−1))\Rightarrow \dfrac{{{\left( 1+i \right)}^{2}}\left( -2+i \right)}{\left( 4-\left( -1 \right) \right)}
⇒(1+i)2(−2+i)5\Rightarrow \dfrac{{{\left( 1+i \right)}^{2}}\left( -2+i \right)}{5}
Now we will expand the (1+i)2{{\left( 1+i \right)}^{2}} in the numerator,
By expanding we will get,
⇒(1+2i−1)(−2+i)5\Rightarrow \dfrac{\left( 1+2i-1 \right)\left( -2+i \right)}{5}
By further simplifying we will get,
⇒2i(−2+i)5\Rightarrow \dfrac{2i\left( -2+i \right)}{5}
Now we have to multiply the terms in the numerator,
By multiplying we will get,
⇒(−2(2i)+2(i)2)5\Rightarrow \dfrac{\left( -2\left( 2i \right)+2{{\left( i \right)}^{2}} \right)}{5}
By further simplifying we will get,
⇒(−4i+2(−1))5\Rightarrow \dfrac{\left( -4i+2\left( -1 \right) \right)}{5}
⇒(−4i−2)5\Rightarrow \dfrac{\left( -4i-2 \right)}{5}
⇒−25−4i5\Rightarrow \dfrac{-2}{5}-\dfrac{4i}{5}
As we know that if a complex number is a+iba+ib then a is said to be the real part and b is said to be an imaginary part.
Therefore, here the imaginary part is −45-\dfrac{4}{5} that is option (D)\left( D \right).

Note: Students should know the concept of complex number, here option (A)45\left( A \right)\dfrac{4}{5} is not the valid answer because the negative sign is missing, we should consider sign also only option (D)−45\left( D \right)-\dfrac{4}{5} is correct. Students should not make any calculation mistakes. Mainly we should know the real and imaginary parts in the function a+iba+ib to solve this question.