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NCERT Exemplar · Q13

Q.If (a2+1)22a−i=x+iy\dfrac{(a^2+1)^2}{2a-i}=x+iy, what is the value of x2+y2x^2+y^2?

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The key idea is that x2+y2x^2+y^2 equals ∣x+iy∣2|x+iy|^2, the squared modulus of the complex number. By taking the modulus of both sides and simplifying, we find x2+y2=(a2+1)44a2+1x^2+y^2 = \frac{(a^2+1)^4}{4a^2+1}.

When you see a complex number expressed as x+iyx+iy, the expression x2+y2x^2+y^2 is not just a random sum — it’s the square of the distance from the origin in the complex plane. That distance is the modulus (or absolute value) of the complex number. So x2+y2=∣x+iy∣2x^2+y^2 = |x+iy|^2.

This is a powerful shortcut: instead of finding xx and yy separately (which would involve rationalising denominators and separating real and imaginary parts), we can work directly with the modulus. The modulus has a beautiful property: for any two complex numbers z1z_1 and z2z_2, ∣z1/z2∣=∣z1∣/∣z2∣|z_1/z_2| = |z_1|/|z_2|, and ∣z1z2∣=∣z1∣∣z2∣|z_1 z_2| = |z_1||z_2|. That’s exactly what we need here.

Let’s apply this.

  1. We are given:

(a2+1)22a−i=x+iy\frac{(a^2+1)^2}{2a - i} = x + iy

Here aa is a real number (since it appears in a real expression a2+1a^2+1 and in the denominator 2a−i2a - i). So a∈Ra \in \mathbb{R}.

  1. Take the modulus of both sides:

∣(a2+1)22a−i∣=∣x+iy∣\left| \frac{(a^2+1)^2}{2a - i} \right| = |x + iy|

  1. Using the property ∣z1/z2∣=∣z1∣/∣z2∣|z_1/z_2| = |z_1|/|z_2|:

∣(a2+1)2∣∣2a−i∣=∣x+iy∣\frac{|(a^2+1)^2|}{|2a - i|} = |x + iy|

  1. The numerator (a2+1)2(a^2+1)^2 is a positive real number, so its modulus is itself:

∣(a2+1)2∣=(a2+1)2|(a^2+1)^2| = (a^2+1)^2

  1. The denominator 2a−i2a - i is a complex number. Its modulus is:

∣2a−i∣=(2a)2+(−1)2=4a2+1|2a - i| = \sqrt{(2a)^2 + (-1)^2} = \sqrt{4a^2 + 1}

  1. Therefore: ∣x+iy∣=(a2+1)24a2+1|x + iy| = \frac{(a^2+1)^2}{\sqrt{4a^2 + 1}} …

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