JEE Main 2022 · 30 Jun · Shift 1 · Mathematics Q2

MathematicsSingle correct+4 / -1Question 2 of 90

The real part of the complex number $\dfrac{(1+2i)^8\cdot(1-2i)^2}{(3+2i)\cdot\overline{(4-6i)}}$ is equal to :

\dfrac{500}{13}

\dfrac{110}{13}

\dfrac{55}6

\dfrac{550}{13}

Answer and worked solution

Official answer: D

Simplify. $(1+2i)^8(1-2i)^2=5^2(1+2i)^6$ , and $(3+2i)(4+6i)=26i$ .

Real part. $(1+2i)^6=(-3+4i)^3=117+44i$ ; dividing $25(117+44i)$ by $26i$ gives real part $\dfrac{25\times44}{26}=\dfrac{550}{13}$ .

(D)

Topics:Algebra of complex numbers

JEE Main 2022 · 30 Jun · Shift 1, Mathematics question 2