Question 20 of 28advanced🔧 ApplyLong Answer4 marks

If a+ib=(x+i)22x2+1a + ib = \frac{(x+i)^2}{2x^2 + 1}, prove that a2+b2=(x2+1)22x2+12a^2 + b^2 = \frac{(x^2 + 1)^2}{2x^2 + 1}^2.

Correct Answer

Given:

a+ib=(x+i)22x2+1a + ib = \frac{(x+i)^2}{2x^2 + 1}

Step 1: Find the modulus of both sides.

We know from the context that for a complex number z=a+ibz = a + ib, the modulus is ∣z∣=a2+b2|z| = \sqrt{a^2 + b^2}.

Taking modulus on both sides:

∣a+ib∣=∣(x+i)22x2+1∣|a + ib| = \left|\frac{(x+i)^2}{2x^2 + 1}\right|

Step 2: Apply the property of modulus.

∣a+ib∣=∣(x+i)2∣∣2x2+1∣=∣x+i∣22x2+1|a + ib| = \frac{|(x+i)^2|}{|2x^2 + 1|} = \frac{|x+i|^2}{2x^2 + 1}

(Since 2x2+1>02x^2 + 1 > 0 for all real xx)

Step 3: Calculate ∣x+i∣|x+i|.

∣x+i∣=x2+1|x+i| = \sqrt{x^2 + 1}

Therefore,

∣a+ib∣=(x2+1)22x2+1=x2+12x2+1|a + ib| = \frac{(\sqrt{x^2 + 1})^2}{2x^2 + 1} = \frac{x^2 + 1}{2x^2 + 1}

Step 4: Square both sides to get a2+b2a^2 + b^2.

Since ∣a+ib∣=a2+b2|a + ib| = \sqrt{a^2 + b^2}, we have:

a2+b2=x2+12x2+1\sqrt{a^2 + b^2} = \frac{x^2 + 1}{2x^2 + 1}

Squaring both sides:

a2+b2=(x2+1)2(2x2+1)2a^2 + b^2 = \frac{(x^2 + 1)^2}{(2x^2 + 1)^2}

∴ Hence proved.

Exercise: Miscellaneous Exercise on Chapter 4 | Q: 6 | (Page 10)
For More Understanding

Explanation

This question tests the concept of modulus of complex numbers. The key insight is that a2+b2=∣z∣2a^2 + b^2 = |z|^2 where z=a+ibz = a + ib. Instead of expanding the numerator (x+i)2(x+i)^2 and separating real and imaginary parts (which is tedious), we use the modulus property ∣z1/z2∣=∣z1∣/∣z2∣|z_1/z_2| = |z_1|/|z_2|.

The denominator 2x2+12x^2 + 1 is always positive for real xx, so its modulus is itself. This approach is demonstrated in Example 8 of the Miscellaneous Examples in the textbook context.

Solution Steps

  1. Step 1: Take modulus on both sides of the given equation

  2. Step 2: Apply the property that modulus of a quotient equals quotient of moduli

  3. Step 3: Calculate ∣x+i∣=x2+1|x+i| = \sqrt{x^2 + 1}

  4. Step 4: Square both sides to obtain a2+b2a^2 + b^2

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