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Complex Numbers question

2022 · 30 Jun · Shift 1 · Q24
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  5. /2022 · 30 Jun · Shift 1 · Q24

Complex Numbers question

2022 · 30 Jun · Shift 1 · Q24

JEE MainMathematicsComplex NumbersMCQ+4 / −1
The real part of the complex number (1+2i)8 . (1−2i)2(3+2i) . (4−6i)‾{{{{(1 + 2i)}^8}\,.\,{{(1 - 2i)}^2}} \over {(3 + 2i)\,.\,\overline {(4 - 6i)} }}(3+2i).(4−6i)​(1+2i)8.(1−2i)2​ is equal to :
  1. A
    50013{{500} \over {13}}13500​
  2. B
    11013{{110} \over {13}}13110​
  3. C
    556{{55} \over {6}}655​
  4. D
    55013{{550} \over {13}}13550​
View written solutionFree

Correct answer: D

  1. Interpret the expression

We need the real part of

(1+2i)8(1−2i)2(3+2i) (4−6i)‾.\frac{(1+2i)^8(1-2i)^2}{(3+2i)\,\overline{(4-6i)}}.(3+2i)(4−6i)​(1+2i)8(1−2i)2​.

Since

(4−6i)‾=4+6i,\overline{(4-6i)}=4+6i,(4−6i)​=4+6i,

the expression becomes

(1+2i)8(1−2i)2(3+2i)(4+6i).\frac{(1+2i)^8(1-2i)^2}{(3+2i)(4+6i)}.(3+2i)(4+6i)(1+2i)8(1−2i)2​.
  1. Simplify the numerator

Note that

(1+2i)(1−2i)=1+4=5.(1+2i)(1-2i)=1+4=5.(1+2i)(1−2i)=1+4=5.

So,

(1+2i)8(1−2i)2=(1+2i)6((1+2i)2(1−2i)2).(1+2i)^8(1-2i)^2=(1+2i)^6\big((1+2i)^2(1-2i)^2\big).(1+2i)8(1−2i)2=(1+2i)6((1+2i)2(1−2i)2).

But

(1+2i)2(1−2i)2=((1+2i)(1−2i))2=52=25.(1+2i)^2(1-2i)^2=\big((1+2i)(1-2i)\big)^2=5^2=25.(1+2i)2(1−2i)2=((1+2i)(1−2i))2=52=25.

Hence,

(1+2i)8(1−2i)2=25(1+2i)6.(1+2i)^8(1-2i)^2=25(1+2i)^6.(1+2i)8(1−2i)2=25(1+2i)6.

Now compute powers of 1+2i1+2i1+2i:

(1+2i)2=1+4i−4=−3+4i.(1+2i)^2=1+4i-4=-3+4i.(1+2i)2=1+4i−4=−3+4i.

Then

(1+2i)4=(−3+4i)2=9−24i−16=−7−24i.(1+2i)^4=(-3+4i)^2=9-24i-16=-7-24i.(1+2i)4=(−3+4i)2=9−24i−16=−7−24i.

And

(1+2i)6=(1+2i)4(1+2i)2=(−7−24i)(−3+4i).(1+2i)^6=(1+2i)^4(1+2i)^2=(-7-24i)(-3+4i).(1+2i)6=(1+2i)4(1+2i)2=(−7−24i)(−3+4i).

Multiply:

(−7)(−3)+(−7)(4i)+(−24i)(−3)+(−24i)(4i)(-7)(-3)+(-7)(4i)+(-24i)(-3)+(-24i)(4i)(−7)(−3)+(−7)(4i)+(−24i)(−3)+(−24i)(4i) =21−28i+72i−96i2=21+44i+96=117+44i.=21-28i+72i-96i^2=21+44i+96=117+44i.=21−28i+72i−96i2=21+44i+96=117+44i.

Thus numerator is

25(117+44i)=2925+1100i.25(117+44i)=2925+1100i.25(117+44i)=2925+1100i.
  1. Simplify the denominator
(3+2i)(4+6i)=12+18i+8i+12i2=12+26i−12=26i.(3+2i)(4+6i)=12+18i+8i+12i^2=12+26i-12=26i.(3+2i)(4+6i)=12+18i+8i+12i2=12+26i−12=26i.

So the given complex number is

2925+1100i26i.\frac{2925+1100i}{26i}.26i2925+1100i​.
  1. Divide by 26i26i26i

Use

1i=−i.\frac{1}{i}=-i.i1​=−i.

Then

2925+1100i26i=126(2925+1100i)⋅1i=126(2925+1100i)(−i).\frac{2925+1100i}{26i}=\frac{1}{26}(2925+1100i)\cdot \frac{1}{i} =\frac{1}{26}(2925+1100i)(-i).26i2925+1100i​=261​(2925+1100i)⋅i1​=261​(2925+1100i)(−i).

Now,

(2925+1100i)(−i)=−2925i−1100i2=1100−2925i.(2925+1100i)(-i)=-2925i-1100i^2=1100-2925i.(2925+1100i)(−i)=−2925i−1100i2=1100−2925i.

Therefore,

2925+1100i26i=110026−292526i=55013−292526i.\frac{2925+1100i}{26i}=\frac{1100}{26}-\frac{2925}{26}i=\frac{550}{13}-\frac{2925}{26}i.26i2925+1100i​=261100​−262925​i=13550​−262925​i.

Hence the real part is

55013.\boxed{\frac{550}{13}}.13550​​.
  1. Check options

The value matches:

D: 55013.\boxed{\text{D: } \frac{550}{13}}.D: 13550​​.
  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.

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