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Question: Find the imaginary part of \(\dfrac{{{\left( 1+i \right)}^{2}}}{\left( 2-i \right)}\) is: a). \(\d...

Find the imaginary part of (1+i)2(2−i)\dfrac{{{\left( 1+i \right)}^{2}}}{\left( 2-i \right)} is:
a). 15\dfrac{1}{5}
b). 35\dfrac{3}{5}
c). 45\dfrac{4}{5}
d). None of these

Explanation

Solution

Hint: We will first simplify the given complex number by expanding the numerator and then multiplying and dividing by the complex conjugate of the denominator. Having simplified it, we will compare it with the general form of complex number a+bia+bi . The value of b is our answer.

Complete step-by-step solution -
Given complex number is (1+i)2(2−i)\dfrac{{{\left( 1+i \right)}^{2}}}{\left( 2-i \right)}
Expanding the numerator using the formula (a+b)2=a2+2ab+b2{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} we get,
(1+i)22−i=i2+2×1×i+i22−i\dfrac{{{\left( 1+i \right)}^{2}}}{2-i}=\dfrac{{{i}^{2}}+2\times 1\times i+{{i}^{2}}}{2-i}
Since i2=−1{{i}^{2}}=-1 ,
(1+i)22−i=1+2i−1(2−i)\dfrac{{{\left( 1+i \right)}^{2}}}{2-i}=\dfrac{1+2i-1}{\left( 2-i \right)} .
Thus, (1+i)22−i=2i2−i\dfrac{{{\left( 1+i \right)}^{2}}}{2-i}=\dfrac{2i}{2-i} .
Now, multiplying and dividing by the complex conjugate of 2-I we get,
(1+i)22−i=2i2−i×2+i2+i................2+i\dfrac{{{\left( 1+i \right)}^{2}}}{2-i}=\dfrac{2i}{2-i}\times \dfrac{2+i}{2+i}................2+i is the complex conjugate of 2-i.
Now multiplying and simplifying, we get,
(1+i)22−i=(2i)×(2+i)(2−i)×(2+i) (1+i)22−i=(2i×2)+(2i×i)(2×2)+(2×i)+(−i×2)+(−i×i) (1+i)22−i=4i+2i24+2i−2i−i2 \begin{aligned} & \dfrac{{{\left( 1+i \right)}^{2}}}{2-i}=\dfrac{\left( 2i \right)\times \left( 2+i \right)}{\left( 2-i \right)\times \left( 2+i \right)} \\\ & \dfrac{{{\left( 1+i \right)}^{2}}}{2-i}=\dfrac{\left( 2i\times 2 \right)+\left( 2i\times i \right)}{\left( 2\times 2 \right)+\left( 2\times i \right)+\left( -i\times 2 \right)+\left( -i\times i \right)} \\\ & \dfrac{{{\left( 1+i \right)}^{2}}}{2-i}=\dfrac{4i+2{{i}^{2}}}{4+2i-2i-{{i}^{2}}} \\\ \end{aligned}
Since i2=−1{{i}^{2}}=-1 , we get.
(1+i)2(2−i)=4i+2(−1)4−(−1) (1+i)2(2−i)=4i−24+1 (1+i)2(2−i)=−2+4i5 (1+i)2(2−i)=−25+4i5 \begin{aligned} & \dfrac{{{\left( 1+i \right)}^{2}}}{\left( 2-i \right)}=\dfrac{4i+2\left( -1 \right)}{4-\left( -1 \right)} \\\ & \dfrac{{{\left( 1+i \right)}^{2}}}{\left( 2-i \right)}=\dfrac{4i-2}{4+1} \\\ & \dfrac{{{\left( 1+i \right)}^{2}}}{\left( 2-i \right)}=\dfrac{-2+4i}{5} \\\ & \dfrac{{{\left( 1+i \right)}^{2}}}{\left( 2-i \right)}=\dfrac{-2}{5}+\dfrac{4i}{5} \\\ \end{aligned}
Now, finally we have,
(1+i)2(2−i)=−25+4i5\dfrac{{{\left( 1+i \right)}^{2}}}{\left( 2-i \right)}=\dfrac{-2}{5}+\dfrac{4i}{5}
Now, comparing this with the general form of the complex number a+bi, where a is the real part and b is the imaginary part, we have.
a+bi=−25+4i5a+bi=\dfrac{-2}{5}+\dfrac{4i}{5}
Thus, a=−25a=\dfrac{-2}{5} and b=45b=\dfrac{4}{5} .
Thus, the imaginary part of the complex number (1+i)2(2−i)\dfrac{{{\left( 1+i \right)}^{2}}}{\left( 2-i \right)} is 45\dfrac{4}{5}
Therefore, option (c) is correct.

Note: We can also solve this problem by converting the complex number to polar form. Thus, the given number is (1+i)22−i\dfrac{{{\left( 1+i \right)}^{2}}}{2-i}.
Now, 1+i1+i has to be converted to polar form, using the formula,
, a+bi=(a2+b2)eiθa+bi=\left( \sqrt{{{a}^{2}}+{{b}^{2}}} \right){{e}^{i\theta }} , where,
(i) θ=tan⁡−1(∣ba∣)\theta ={{\tan }^{-1}}\left( \left| \dfrac{b}{a} \right| \right) , if (a,b)∈I\left( a,b \right)\in I quadrant.
(ii) θ=π−tan⁡−1(∣ba∣)\theta =\pi -{{\tan }^{-1}}\left( \left| \dfrac{b}{a} \right| \right) , if (a,b)∈II\left( a,b \right)\in II quadrant.
(iii) θ=π+tan⁡−1(∣ba∣)\theta =\pi +{{\tan }^{-1}}\left( \left| \dfrac{b}{a} \right| \right) , if (a,b)∈III\left( a,b \right)\in III quadrant.
(iv) θ=−tan⁡−1(∣ba∣)\theta =-{{\tan }^{-1}}\left( \left| \dfrac{b}{a} \right| \right) , if (a,b)∈IV\left( a,b \right)\in IV quadrant.
Now, for 1+i1+i , a=1 and b=1.
∴(a,b) that is (1,1) belongs to I quadrant\therefore \left( a,b \right)\text{ that is }\left( 1,1 \right)\text{ belongs to I quadrant}
Thus 1+i=(12+12)eiθ1+i=\left( \sqrt{{{1}^{2}}+{{1}^{2}}} \right){{e}^{i\theta }} , where θ=tan⁡−1(∣11∣)\theta ={{\tan }^{-1}}\left( \left| \dfrac{1}{1} \right| \right) .
⇒1+i=2eiπ4\Rightarrow 1+i=\sqrt{2}{{e}^{\dfrac{i\pi }{4}}} , Since θ=tan⁡−1(1)\theta ={{\tan }^{-1}}\left( 1 \right)
θ=π4\theta =\dfrac{\pi }{4}
Now, similarly we convert 2−i2-i to polar form.
Here, a=2, b=-1
Therefore, (a,b)∈IV\left( a,b \right)\in IV Quadrant
Thus, 2−i=(22+(−1)2)eiθ2-i=\left( \sqrt{{{2}^{2}}+{{\left( -1 \right)}^{2}}} \right){{e}^{i\theta }} , θ=−tan⁡−1(∣−12∣)\theta =-{{\tan }^{-1}}\left( \left| \dfrac{-1}{2} \right| \right).
⇒2−i=(4+1)eiθ\Rightarrow 2-i=\left( \sqrt{4+1} \right){{e}^{i\theta }} , θ=−tan⁡−1(12)\theta =-{{\tan }^{-1}}\left( \dfrac{1}{2} \right) .
⇒2−i=5eitan⁡−1(12)\Rightarrow 2-i=\sqrt{5}{{e}^{i{{\tan }^{-1}}\left( \dfrac{1}{2} \right)}}
Now, putting these polar forms in place, we get,
(1+i)22−i=(2eiπ4)25e−itan⁡−1(12)\dfrac{{{\left( 1+i \right)}^{2}}}{2-i}=\dfrac{{{\left( \sqrt{2}{{e}^{\dfrac{i\pi }{4}}} \right)}^{2}}}{\sqrt{5}{{e}^{-i{{\tan }^{-1}}\left( \dfrac{1}{2} \right)}}}
(1+i)22−i=(2)2(eiπ4×2)5e−itan⁡−1(12)\dfrac{{{\left( 1+i \right)}^{2}}}{2-i}=\dfrac{{{\left( \sqrt{2} \right)}^{2}}\left( {{e}^{\dfrac{i\pi }{4}\times 2}} \right)}{\sqrt{5}{{e}^{-i{{\tan }^{-1}}\left( \dfrac{1}{2} \right)}}} ……………… using (am)n=amn{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}} .
(1+i)22−i=2eiπ25e−itan⁡−1(12)\dfrac{{{\left( 1+i \right)}^{2}}}{2-i}=\dfrac{2{{e}^{\dfrac{i\pi }{2}}}}{\sqrt{5}{{e}^{-i{{\tan }^{-1}}\left( \dfrac{1}{2} \right)}}}
(1+i)22−i=2eiπ2+itan⁡−1(12)5\dfrac{{{\left( 1+i \right)}^{2}}}{2-i}=\dfrac{2{{e}^{\dfrac{i\pi }{2}+i{{\tan }^{-1}}\left( \dfrac{1}{2} \right)}}}{\sqrt{5}} ………………………………. Using ama−n=am+n\dfrac{{{a}^{m}}}{{{a}^{-n}}}={{a}^{m+n}} .
(1+i)22−i=2ei(π2+tan⁡−1(12))5\dfrac{{{\left( 1+i \right)}^{2}}}{2-i}=\dfrac{2{{e}^{i\left( ^{\dfrac{\pi }{2}+{{\tan }^{-1}}\left( \dfrac{1}{2} \right)} \right)}}}{\sqrt{5}}
Now, we know, Euler’s formula,
eiθ=cos⁡θ+isin⁡θ{{e}^{i\theta }}=\cos \theta +i\sin \theta , and we also know that
cos⁡(π2+θ)=−sin⁡θ\cos \left( \dfrac{\pi }{2}+\theta \right)=-\sin \theta and sin⁡(π2+θ)=cos⁡θ\sin \left( \dfrac{\pi }{2}+\theta \right)=\cos \theta .
Now, putting θ\theta as π2+tan⁡−1(12)\dfrac{\pi }{2}+{{\tan }^{-1}}\left( \dfrac{1}{2} \right) we get,
ei(π2+tan⁡−1(12))=cos⁡(π2+tan⁡−1(12))+isin⁡(π2+tan⁡−1(12)) =−sin⁡(tan⁡−1(12))+icos⁡(tan⁡−1(12)) \begin{aligned} & {{e}^{i\left( \dfrac{\pi }{2}+{{\tan }^{-1}}\left( \dfrac{1}{2} \right) \right)}}=\cos \left( \dfrac{\pi }{2}+{{\tan }^{-1}}\left( \dfrac{1}{2} \right) \right)+i\sin \left( \dfrac{\pi }{2}+{{\tan }^{-1}}\left( \dfrac{1}{2} \right) \right) \\\ & =-\sin \left( {{\tan }^{-1}}\left( \dfrac{1}{2} \right) \right)+i\cos \left( {{\tan }^{-1}}\left( \dfrac{1}{2} \right) \right) \\\ \end{aligned}
Now, let α=tan⁡−1(12)\alpha ={{\tan }^{-1}}\left( \dfrac{1}{2} \right)
Therefore, tan⁡α=(12)\tan \alpha =\left( \dfrac{1}{2} \right)
Now, tan⁡α=opposite sideAdjacent side\tan \alpha =\dfrac{\text{opposite side}}{\text{Adjacent side}} .
Thus, opposite side = 1 and adjacent side = 2.
Therefore, hypotenuse =(opposite side)2+(Adjacent side)2=\sqrt{{{\left( opposite\text{ side} \right)}^{2}}+{{\left( \text{Adjacent side} \right)}^{2}}}

& hypotenuse=\sqrt{{{1}^{2}}+{{2}^{2}}} \\\ & hypotenuse=\sqrt{1+4} \\\ & hypotenuse=\sqrt{5} \\\ \end{aligned}$$ Thus, $\sin \alpha =\dfrac{\text{opposite side}}{hypotenuse}$ $\sin \alpha =\dfrac{1}{\sqrt{5}}$ Applying ${{\sin }^{-1}}$ on both sides, we get, $\alpha ={{\sin }^{-1}}\left( \dfrac{1}{\sqrt{5}} \right)$ Similarly $\cos \alpha =\dfrac{\text{Adjacent side}}{hypotenuse}$ Which is $\cos \alpha =\dfrac{2}{\sqrt{5}}$ Applying ${{\cos }^{-1}}$ on both sides, $\alpha ={{\cos }^{-1}}\left( \dfrac{2}{\sqrt{5}} \right)$ Thus, ${{\tan }^{-1}}\left( \dfrac{1}{2} \right)={{\sin }^{-1}}\left( \dfrac{1}{\sqrt{5}} \right)$ And ${{\tan }^{-1}}\left( \dfrac{1}{2} \right)={{\cos }^{-1}}\left( \dfrac{2}{\sqrt{5}} \right)$ . Thus in the Euler’s formula, $\begin{aligned} & {{e}^{i\left( \dfrac{\pi }{2}+{{\tan }^{-1}}\left( \dfrac{1}{2} \right) \right)}}=-\sin \left( {{\tan }^{-1}}\left( \dfrac{1}{2} \right) \right)+i\cos \left( {{\tan }^{-1}}\left( \dfrac{1}{2} \right) \right) \\\ & =-\sin \left( {{\sin }^{-1}}\left( \dfrac{1}{\sqrt{5}} \right) \right)+i\cos \left( {{\cos }^{-1}}\left( \dfrac{2}{\sqrt{5}} \right) \right) \\\ \end{aligned}$ Now, sin and ${{\sin }^{-1}}$ cancel out each other. Similarly cos and ${{\cos }^{-1}}$ cancel out of each other. Thus, ${{e}^{i\left( \dfrac{\pi }{2}+{{\tan }^{-1}}\left( \dfrac{1}{2} \right) \right)}}=-\dfrac{1}{\sqrt{5}}+\dfrac{i2}{\sqrt{5}}$ Thus, $\dfrac{{{\left( 1+i \right)}^{2}}}{2-i}=\dfrac{2}{\sqrt{5}}{{e}^{i\left( \dfrac{\pi }{2}+{{\tan }^{-1}}\left( \dfrac{1}{2} \right) \right)}}$ . $\Rightarrow \dfrac{{{\left( 1+i \right)}^{2}}}{2-i}=\dfrac{2}{\sqrt{5}}\left( \dfrac{-1}{\sqrt{5}}+\dfrac{2}{\sqrt{5}}i \right)$ $$\begin{aligned} & \Rightarrow \dfrac{{{\left( 1+i \right)}^{2}}}{2-i}=\dfrac{-2}{{{\left( \sqrt{5} \right)}^{2}}}+\dfrac{2\times 2}{{{\left( \sqrt{5} \right)}^{2}}}i \\\ & \Rightarrow \dfrac{{{\left( 1+i \right)}^{2}}}{2-i}=\dfrac{-2}{5}+\dfrac{4i}{5} \\\ \end{aligned}$$ Thus, the complex number has $\left( \dfrac{4}{5} \right)$ as the imaginary part.