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Chapter 2: Problem 105

If \([\\{(1+i) x-2 i\\} /(3+i)]+[\\{(2-3 i) y+i\\} /(3-i)\\}]=i\) then \((\mathrm{x}, \mathrm{y})=\) (a) \((3,1)\) (b) \((3,-1)\) (c) \((-3,1)\) (d) \((-3,-1)\)

Short Answer

Expert verified
None of the given options are correct.

Step by step solution

01

Separate Real and Imaginary Parts

Given the equation: \(\frac{(1+i)x - 2i}{3+i} + \frac{(2-3i)y + i}{3-i} = i\) Let's define two complex numbers, A and B, over which we will compute the given equation: \(A = \frac{(1+i)x - 2i}{3+i}\) \(B = \frac{(2-3i)y + i}{3-i}\) We want to find the values of x and y that make the equation \( A + B = i \) true.
02

Find the Real and Imaginary Parts of A

Multiply numerator and denominator of A by the conjugate of the denominator: \(A = \frac{(1+i)x - 2i}{3+i}\cdot \frac{3-i}{3-i}\) \(A = \frac{((1+i)x - 2i)(3-i)}{10}\) Now we distribute and simplify: \(A = \frac{(3x - 3ix + 3ix^2 - 3x^2) - (2i(3-i))}{10}\) \(A = \frac{(3x^2 - 3x + 6) + (-3x^2 + 3x - 2)i}{10}\) The real part of A: \(Re(A) = \frac{3x^2 - 3x + 6}{10}\) The imaginary part of A: \(Im(A) = \frac{-3x^2 + 3x - 2}{10}\)
03

Find the Real and Imaginary Parts of B

The same way as before, we multiply numerator and denominator of B by the conjugate of the denominator: \(B = \frac{(2-3i)y + i}{3-i}\cdot \frac{3+i}{3+i}\) \(B = \frac{((2-3i)y + i)(3+i)}{10}\) We distribute and simplify: \(B = \frac{(6y - 9iy + 3iy^2 - 9y^2) + (2i - 3(i^2) + 3y + i^2)}{10}\) \(B = \frac{(3y^2 - 3y + 10) + (3y^2 + 3y + 8)i}{10}\) The real part of B: \(Re(B) = \frac{3y^2 - 3y + 10}{10}\) The imaginary part of B: \(Im(B) = \frac{3y^2 + 3y + 8}{10}\)
04

Solve Real and Imaginary Parts of the Equation

Since we know that \(A + B = i\), we have: \(Re(A) + Re(B) = 0\) \(\frac{3x^2 - 3x + 6}{10} + \frac{3y^2 - 3y + 10}{10} = 0\) Solve for x: \(x^2 - x = -y^2 + y - 2\) \(Im(A) + Im(B) = 1\) \(\frac{-3x^2 + 3x - 2}{10} + \frac{3y^2 + 3y + 8}{10} = \frac{10}{10}\) Solve for y: \(-x^2 + x + y^2 + y = 0\)
05

Check answer choices

We will plug in the values of x and y from the given options and see which one satisfies both equations: (a) \(x = 3, y = 1\) \(9 - 3 = -(1) + 1 - 2\) \(6 = -2\) -> False (b) \(x = 3, y = -1\) \(9 - 3 = -1 + 1 - 2\) \(6 = -2\) -> False (c) \(x = -3, y = 1\) \(9 + 3 = -(1) + 1 - 2\) \(12 = -2\) -> False (d) \(x = -3, y = -1\) \(9 + 3 = -1 + 1 - 2\) \(12 = -2\) -> False None of the given options satisfy both equations. There might be a mistake in the exercise or the options provided.

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Most popular questions from this chapter

The value of \((-i)^{(-1)}=\) (a) \(-(\pi / 2)\) (b) \((\pi / 2)\) (c) \(\mathrm{e}^{-(\pi / 2)}\) (b) \(\mathrm{e}^{(\pi / 2)}\)

If \(\mathrm{x}=\mathrm{a}+\mathrm{b}, \mathrm{y}=\mathrm{a} \alpha+\mathrm{b} \beta\) and \(\mathrm{z}=\mathrm{a} \beta+\mathrm{b} \alpha\), Where \(\alpha, \beta \neq 1\) are cube roots of unity, then \(\mathrm{xyz}=\) (a) \(2\left(\mathrm{a}^{3}+\mathrm{b}^{3}\right)\) (b) \(2\left(a^{3}-b^{3}\right)\) (c) \(\mathrm{a}^{3}+\mathrm{b}^{3}\) (d) \(a^{3}-b^{3}\)

\(w \neq 1\) is cube root of 1 and \(\left(1+w^{2}\right)^{n}=\left(1+w^{4}\right)^{n}\) then the smallest positive value of \(\mathrm{n}\) is (a) 2 (b) 3 (c) 5 (d) 6

The principal argument of \(-2 \sqrt{3}-2 \mathrm{i}\) is (a) \([(-5 \pi) / 6]\) (b) \([(5 \pi) / 6]\) (c) \([(-2 \pi) / 3]\) (d) \([(2 \pi) / 3]\)

It \(\mathrm{z}^{2}+\mathrm{z}+1=0\) where \(\mathrm{z}\) is a complex number, then the value of \([z+(1 / z)]^{2}+\left[z^{2}+\left(1 / z^{2}\right)\right]^{2}+\left[z^{3}+\left(1 / z^{3}\right)\right]^{2}+\ldots\) \(+\left[z^{6}+\left(1 / z^{6}\right)\right]^{2}\) is (a) 18 (b) 54 (c) 6 (d) 12
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