Question 22 of 28intermediate🔧 ApplyNumerical2 marks

Find the real numbers xx and yy if (x−iy)(3+5i)(x - iy) (3 + 5i) is the conjugate of −6−24i-6 - 24i.

Correct Answer

x = 3 and y = -3

Exercise: Miscellaneous Exercise on Chapter 4 | Q: 8 | (Page 10)
For More Understanding

Explanation

This question tests the concept of complex conjugates and equating real and imaginary parts. The conjugate of a complex number a + bi is a - bi. After finding the conjugate, we expand the product and equate real and imaginary parts to solve for x and y.

Solution Steps

  1. Step 1: Find the conjugate of −6−24i-6 - 24i. The conjugate is −6+24i-6 + 24i.

  2. Step 2: Set up the equation: (x−iy)(3+5i)=−6+24i(x - iy)(3 + 5i) = -6 + 24i

  3. Step 3: Expand the left side: (x−iy)(3+5i)=3x+5xi−3iy−5i2y=3x+5y+i(5x−3y)(x - iy)(3 + 5i) = 3x + 5xi - 3iy - 5i^2y = 3x + 5y + i(5x - 3y)

  4. Step 4: Equate real and imaginary parts: 3x+5y=−63x + 5y = -6 and 5x−3y=245x - 3y = 24

  5. Step 5: Solve the system of equations. Multiplying first equation by 3: 9x+15y=−189x + 15y = -18. Multiplying second equation by 5: 25x−15y=12025x - 15y = 120. Adding: 34x=10234x = 102, so x=3x = 3.

  6. Step 6: Substitute x=3x = 3 into 3x+5y=−63x + 5y = -6: 9+5y=−69 + 5y = -6, giving 5y=−155y = -15, so y=−3y = -3.

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