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JEE Main 2021
Complex Numbers
Complex Numbers
Easy

Question

Let (213i)3=x+iy27(i=1),{\left( { - 2 - {1 \over 3}i} \right)^3} = {{x + iy} \over {27}}\left( {i = \sqrt { - 1} } \right),\,\, where x and y are real numbers, then y - x equals :

Options

Solution

Key Concepts and Formulas

  • Binomial Expansion: (a+b)3=a3+3a2b+3ab2+b3(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
  • Imaginary Unit Properties: i2=1i^2 = -1, i3=ii^3 = -i
  • Complex Number Equality: If a+bi=c+dia+bi = c+di, where a,b,c,da, b, c, d are real numbers, then a=ca=c and b=db=d.

Step-by-Step Solution

1. Simplify the Complex Number We are given the complex number (213i)3{\left( { - 2 - {1 \over 3}i} \right)^3}. We want to simplify the expression inside the parentheses by finding a common denominator. 213i=6313i=6+i3-2 - \frac{1}{3}i = -\frac{6}{3} - \frac{1}{3}i = -\frac{6+i}{3} This simplifies the base of the expression, making the cubing process easier.

2. Cube the Simplified Complex Number Now, we cube the simplified expression: (213i)3=(6+i3)3=(6+i)333=(6+i)327{\left( { - 2 - {1 \over 3}i} \right)^3} = {\left( {-\frac{6+i}{3}} \right)^3} = -\frac{(6+i)^3}{3^3} = -\frac{(6+i)^3}{27} The negative sign comes out because (a)3=a3(-a)^3 = -a^3.

3. Expand (6+i)3(6+i)^3 using the Binomial Theorem We use the binomial expansion formula (a+b)3=a3+3a2b+3ab2+b3(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 with a=6a=6 and b=ib=i: (6+i)3=63+3(62)(i)+3(6)(i2)+i3(6+i)^3 = 6^3 + 3(6^2)(i) + 3(6)(i^2) + i^3 This expands the complex number raised to the third power.

4. Evaluate Powers of ii and Simplify We substitute i2=1i^2 = -1 and i3=ii^3 = -i into the expanded expression: (6+i)3=216+3(36)i+3(6)(1)+(i)=216+108i18i(6+i)^3 = 216 + 3(36)i + 3(6)(-1) + (-i) = 216 + 108i - 18 - i Now, we group the real and imaginary terms: (6+i)3=(21618)+(1081)i=198+107i(6+i)^3 = (216 - 18) + (108 - 1)i = 198 + 107i Consolidating the terms into the standard form A+BiA+Bi is crucial for the next step.

5. Substitute Back and Compare with the Given Equation We found that (213i)3=198+107i27{\left( { - 2 - {1 \over 3}i} \right)^3} = - \frac{{198 + 107i}}{{27}}. Substituting back into the original equation: (213i)3=198+107i27=198107i27{\left( { - 2 - {1 \over 3}i} \right)^3} = -\frac{198 + 107i}{27} = \frac{-198 - 107i}{27} We are given that (213i)3=x+iy27{\left( { - 2 - {1 \over 3}i} \right)^3} = {{x + iy} \over {27}}. Comparing the two expressions, we can equate the numerators: x+iy=198107ix + iy = -198 - 107i Since xx and yy are real numbers, we can identify them: x=198x = -198 y=107y = -107 By expressing the left-hand side in the same format as the right-hand side, we can directly equate the real and imaginary components.

6. Calculate yxy - x Finally, we substitute the values of xx and yy into the expression yxy - x: yx=107(198)=107+198=91y - x = -107 - (-198) = -107 + 198 = 91 This gives us the final answer.

Common Mistakes & Tips

  • Sign Errors: Pay close attention to signs, especially when cubing negative numbers and distributing negative signs.
  • Powers of ii: Remember the cycle of powers of ii: i,1,i,1i, -1, -i, 1.
  • Binomial Expansion: Double-check the coefficients and powers in the binomial expansion.

Summary We simplified the complex number, applied the binomial expansion, and equated the real and imaginary parts to find xx and yy. Finally, we calculated yxy-x to get 9191.

The final answer is \boxed{91}, which corresponds to option (D).

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