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Complex Numbers question

2019 · 11 Jan · Shift 1 · Q38
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Complex Numbers question

2019 · 11 Jan · Shift 1 · Q38

JEE MainMathematicsComplex NumbersMCQ+4 / −1
Let (−2−13i)3=x+iy27(i=−1),  {\left( { - 2 - {1 \over 3}i} \right)^3} = {{x + iy} \over {27}}\left( {i = \sqrt { - 1} } \right),\,\,(−2−31​i)3=27x+iy​(i=−1​), where x and y are real numbers, then y −-− x equals :
  1. A
    −-− 85
  2. B
    85
  3. C
    −-− 91
  4. D
    91
View written solutionFree

Correct answer: D

  1. Given expression

We need to compute

(−2−13i)3=x+iy27.\left(-2-\frac{1}{3}i\right)^3 = \frac{x+iy}{27}.(−2−31​i)3=27x+iy​.

Then find y−xy-xy−x.

  1. Remove the fraction first

Write

−2−13i=−6−i3.-2-\frac13 i = \frac{-6-i}{3}.−2−31​i=3−6−i​.

So,

(−2−13i)3=(−6−i3)3=(−6−i)327.\left(-2-\frac13 i\right)^3 = \left(\frac{-6-i}{3}\right)^3 = \frac{(-6-i)^3}{27}.(−2−31​i)3=(3−6−i​)3=27(−6−i)3​.

Comparing with

x+iy27,\frac{x+iy}{27},27x+iy​,

we get

x+iy=(−6−i)3.x+iy = (-6-i)^3.x+iy=(−6−i)3.
  1. Compute (−6−i)2(-6-i)^2(−6−i)2
(−6−i)2=36+12i+i2=36+12i−1=35+12i.(-6-i)^2 = 36 + 12i + i^2 = 36+12i-1 = 35+12i.(−6−i)2=36+12i+i2=36+12i−1=35+12i.
  1. Now multiply once more by (−6−i)(-6-i)(−6−i)
(−6−i)3=(35+12i)(−6−i).(-6-i)^3 = (35+12i)(-6-i).(−6−i)3=(35+12i)(−6−i).

Expand:

(35+12i)(−6−i)=35(−6)+35(−i)+12i(−6)+12i(−i).(35+12i)(-6-i)=35(-6)+35(-i)+12i(-6)+12i(-i).(35+12i)(−6−i)=35(−6)+35(−i)+12i(−6)+12i(−i).

So,

=−210−35i−72i−12i2.= -210 - 35i - 72i - 12i^2.=−210−35i−72i−12i2.

Since i2=−1i^2=-1i2=−1,

−12i2=12.-12i^2 = 12.−12i2=12.

Hence,

(−6−i)3=−210+12−107i=−198−107i.(-6-i)^3 = -210 + 12 - 107i = -198 - 107i.(−6−i)3=−210+12−107i=−198−107i.

Therefore,

x=−198,y=−107.x=-198,\qquad y=-107.x=−198,y=−107.
  1. Find y−xy-xy−x
y−x=(−107)−(−198)=91.y-x = (-107)-(-198)=91.y−x=(−107)−(−198)=91.
  1. Check options

The correct option is

91\boxed{91}91​

which is Option D.

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