Skip to content
Miscellaneous Exercise · Q12

Q.Find the number of non-zero integral solutions of the equation ∣1−i∣x=2x|1 - i|^{x} = 2^{x}.

Yanam BieapTextbookSubjective· 2mImportance★★★★★est
39% · 34/88 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The equation ∣1−i∣x=2x|1 - i|^x = 2^x simplifies to (2)x=2x(\sqrt{2})^x = 2^x, which holds only when x=0x = 0. Since we need non-zero integral solutions, there are zero such solutions.

The key insight here is recognizing what ∣1−i∣|1 - i| represents geometrically and algebraically. The modulus of a complex number a+bia + bi is a2+b2\sqrt{a^2 + b^2}, which gives us the distance from the origin in the complex plane.

For 1−i1 - i, we have:

∣1−i∣=12+(−1)2=1+1=2|1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}

Now our equation becomes much clearer. We're looking for integer values of xx (excluding zero) that satisfy:

(2)x=2x(\sqrt{2})^x = 2^x

Let me rewrite both sides using the same base to see what's really happening.

Step-by-step analysis:

  1. Express both sides with base 2.

    Since 2=21/2\sqrt{2} = 2^{1/2}, the left side becomes:

(2)x=(21/2)x=2x/2(\sqrt{2})^x = (2^{1/2})^x = 2^{x/2}

So our equation is:

2x/2=2x2^{x/2} = 2^x

  1. Use the fact that exponential functions are one-to-one.

    Since f(t)=2tf(t) = 2^t is a strictly increasing function (and therefore injective), we have 2a=2b2^a = 2^b if and only if a=ba = b.

    This means:

x2=x\frac{x}{2} = x

  1. Solve for xx.

x2=x\frac{x}{2} = x

x=2xx = 2x

0=x0 = x …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.