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Mathematics Question on Algebra of Complex Numbers

The imaginary part of (1+i)2i(2i−1)\frac{\left(1+i\right)^{2}}{i\left(2i-1\right)} is

A

45\frac{4}{5}

B

0

C

25\frac{2}{5}

D

−45 - \frac{4}{5}

Answer

−45 - \frac{4}{5}

Explanation

Solution

(1+i)2i(2i−1)=1+i2+2i2i2−i\frac{\left(1+i\right)^{2}}{i\left(2i-1\right)} = \frac{1+i^{2}+2i}{2i^{2} -i}
⇒1−1+2i−2−i=−2i2+i×(2−i)(2−i)=−4i+2i24−i2[∵i2=−1]\Rightarrow \frac{1-1+2i}{-2-i} = - \frac{2i}{2+i}\times \frac{\left(2-i\right)}{\left(2-i\right)} = \frac{-4i +2i^{2}}{4-i^{2}} \left[\because i^{2 } = -1\right]
=−4i−24+1=−2−4i5⇒−25−4i5= \frac{-4i-2}{4+1} = \frac{-2-4i}{5} \Rightarrow \frac{-2}{5} - \frac{4i}{5}
∴\therefore The imaginary part =−45= - \frac{4}{5}