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JEE Main 2024
Complex Numbers
Complex Numbers
Hard

Question

If α\alpha denotes the number of solutions of 1ix=2x|1-i|^x=2^x and β=(zarg(z))\beta=\left(\frac{|z|}{\arg (z)}\right), where z=π4(1+i)4[1πiπ+i+πi1+πi],i=1z=\frac{\pi}{4}(1+i)^4\left[\frac{1-\sqrt{\pi} i}{\sqrt{\pi}+i}+\frac{\sqrt{\pi}-i}{1+\sqrt{\pi} i}\right], i=\sqrt{-1}, then the distance of the point (α,β)(\alpha, \beta) from the line 4x3y=74 x-3 y=7 is __________.

Answer: 2

Solution

Key Concepts and Formulas

  • Modulus of a Complex Number: For a complex number z=x+iyz = x + iy, its modulus is z=x2+y2|z| = \sqrt{x^2 + y^2}.
  • Argument of a Complex Number: The argument, arg(z)\arg(z), is the angle zz makes with the positive real axis. For z=iyz=iy with y>0y>0, arg(z)=π2\arg(z) = \frac{\pi}{2}.
  • Distance of a Point from a Line: The distance DD of a point (x1,y1)(x_1, y_1) from a line Ax+By+C=0Ax + By + C = 0 is D=Ax1+By1+CA2+B2D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}.

Step-by-Step Solution

Part 1: Finding the value of α\alpha

Step 1: Calculate the modulus of 1i1-i

We need to find 1i|1-i| to simplify the given equation. 1i=12+(1)2=2|1-i| = \sqrt{1^2 + (-1)^2} = \sqrt{2} Explanation: We use the definition of the modulus of a complex number.

Step 2: Substitute and simplify the equation

Substitute 1i=2|1-i| = \sqrt{2} into the equation 1ix=2x|1-i|^x = 2^x. (2)x=2x(\sqrt{2})^x = 2^x Rewrite 2\sqrt{2} as 21/22^{1/2}: (21/2)x=2x(2^{1/2})^x = 2^x 2x/2=2x2^{x/2} = 2^x Explanation: Expressing both sides with the same base allows us to compare exponents.

Step 3: Solve for xx

Since the bases are equal, the exponents must be equal: x2=x\frac{x}{2} = x x=2xx = 2x x2x=0x - 2x = 0 x=0-x = 0 x=0x = 0 Therefore, α=1\alpha = 1. Explanation: Solving the equation for xx gives the only solution.

Part 2: Finding the value of β\beta

Step 1: Simplify (1+i)4(1+i)^4

We simplify this term first to reduce the complexity of the expression for zz. (1+i)2=1+2i+i2=1+2i1=2i(1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i (1+i)4=((1+i)2)2=(2i)2=4i2=4(1+i)^4 = ((1+i)^2)^2 = (2i)^2 = 4i^2 = -4 Explanation: We use the binomial expansion and the property i2=1i^2 = -1.

Step 2: Simplify 1πiπ+i+πi1+πi\frac{1-\sqrt{\pi} i}{\sqrt{\pi}+i}+\frac{\sqrt{\pi}-i}{1+\sqrt{\pi} i}

We simplify each fraction separately and then add them. 1πiπ+i=(1πi)(πi)(π+i)(πi)=πiπiππ+1=iπiπ+1=i(1+π)1+π=i\frac{1-\sqrt{\pi} i}{\sqrt{\pi}+i} = \frac{(1-\sqrt{\pi} i)(\sqrt{\pi}-i)}{(\sqrt{\pi}+i)(\sqrt{\pi}-i)} = \frac{\sqrt{\pi} - i - \pi i - \sqrt{\pi}}{\pi + 1} = \frac{-i - \pi i}{\pi + 1} = \frac{-i(1+\pi)}{1+\pi} = -i πi1+πi=(πi)(1πi)(1+πi)(1πi)=ππiiπ1+π=iπi1+π=i(1+π)1+π=i\frac{\sqrt{\pi}-i}{1+\sqrt{\pi} i} = \frac{(\sqrt{\pi}-i)(1-\sqrt{\pi}i)}{(1+\sqrt{\pi} i)(1-\sqrt{\pi}i)} = \frac{\sqrt{\pi} - \pi i - i - \sqrt{\pi}}{1+\pi} = \frac{-i - \pi i}{1+\pi} = \frac{-i(1+\pi)}{1+\pi} = -i Therefore, 1πiπ+i+πi1+πi=i+(i)=2i\frac{1-\sqrt{\pi} i}{\sqrt{\pi}+i}+\frac{\sqrt{\pi}-i}{1+\sqrt{\pi} i} = -i + (-i) = -2i Explanation: We multiply the numerator and denominator of each fraction by the conjugate of the denominator to rationalize it.

Step 3: Simplify the expression for zz

z=π4(1+i)4[1πiπ+i+πi1+πi]z = \frac{\pi}{4}(1+i)^4\left[\frac{1-\sqrt{\pi} i}{\sqrt{\pi}+i}+\frac{\sqrt{\pi}-i}{1+\sqrt{\pi} i}\right] z=π4(4)(2i)=2πiz = \frac{\pi}{4}(-4)(-2i) = 2\pi i Explanation: We substitute the simplified values from the previous steps.

Step 4: Calculate z|z| and arg(z)\arg(z)

z=2πi=02+(2π)2=2π|z| = |2\pi i| = \sqrt{0^2 + (2\pi)^2} = 2\pi arg(z)=π2\arg(z) = \frac{\pi}{2} Explanation: We use the definitions of modulus and argument for a purely imaginary number.

Step 5: Calculate β\beta

β=zarg(z)=2ππ2=2π2π=4\beta = \frac{|z|}{\arg(z)} = \frac{2\pi}{\frac{\pi}{2}} = \frac{2\pi \cdot 2}{\pi} = 4 Explanation: We substitute the values of z|z| and arg(z)\arg(z) into the definition of β\beta.

Part 3: Calculate the Distance

Step 1: State the point and the line equation

We have the point (α,β)=(1,4)(\alpha, \beta) = (1, 4) and the line 4x3y=74x - 3y = 7, which can be written as 4x3y7=04x - 3y - 7 = 0.

Step 2: Apply the distance formula

D=4(1)3(4)742+(3)2=412716+9=1525=155=3D = \frac{|4(1) - 3(4) - 7|}{\sqrt{4^2 + (-3)^2}} = \frac{|4 - 12 - 7|}{\sqrt{16 + 9}} = \frac{|-15|}{\sqrt{25}} = \frac{15}{5} = 3 Explanation: We substitute the values into the distance formula and calculate the result.

Common Mistakes & Tips

  • Careless arithmetic, especially with signs, can lead to errors. Double-check each calculation.
  • When simplifying complex fractions, multiplying by the conjugate is a reliable method.
  • Remember the correct formula for the distance between a point and a line.

Summary

By carefully simplifying the given expressions and applying the relevant formulas, we found that α=1\alpha = 1, β=4\beta = 4, and the distance of the point (α,β)(\alpha, \beta) from the line 4x3y=74x - 3y = 7 is 33. There seems to be an error in the question, as the correct distance is 3, not 2 as stated in the provided answer.

Final Answer

The final answer is 3\boxed{3}.

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