Question:
The number of integer solutions of the equation
\[
|1-i|^x=2^x
\]
is
The number of integer solutions of the equation
\[ |1-i|^x=2^x \] is
\[ |1-i|^x=2^x \] is
Show Hint
Always simplify the modulus of a complex number first. Then convert radicals into exponential form for easy comparison of powers.
Updated On: Jun 15, 2026
- \(1\)
- \(0\)
- \(2\)
- \(3\)
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The Correct Option is A
Solution and Explanation
Step 1: Find the modulus of \(1-i\).
We know that for a complex number \(a+ib\),
\[ |a+ib|=\sqrt{a^2+b^2} \]
Therefore,
\[ |1-i|=\sqrt{1^2+(-1)^2} \] \[ =\sqrt{1+1} \] \[ =\sqrt{2} \]
Hence, the equation becomes
\[ (\sqrt2)^x=2^x \]
Step 2: Rewrite in exponential form.
Since
\[ \sqrt2=2^{1/2}, \] we get
\[ (2^{1/2})^x=2^x \]
\[ 2^{x/2}=2^x \]
Since the bases are equal and positive, compare exponents:
\[ \frac{x}{2}=x \]
\[ x=0 \]
Step 3: Count integer solutions.
The only integer solution is
\[ x=0 \]
Therefore, the number of integer solutions is
\[ 1 \]
Step 4: Final conclusion.
Hence,
\[ \boxed{1} \]
We know that for a complex number \(a+ib\),
\[ |a+ib|=\sqrt{a^2+b^2} \]
Therefore,
\[ |1-i|=\sqrt{1^2+(-1)^2} \] \[ =\sqrt{1+1} \] \[ =\sqrt{2} \]
Hence, the equation becomes
\[ (\sqrt2)^x=2^x \]
Step 2: Rewrite in exponential form.
Since
\[ \sqrt2=2^{1/2}, \] we get
\[ (2^{1/2})^x=2^x \]
\[ 2^{x/2}=2^x \]
Since the bases are equal and positive, compare exponents:
\[ \frac{x}{2}=x \]
\[ x=0 \]
Step 3: Count integer solutions.
The only integer solution is
\[ x=0 \]
Therefore, the number of integer solutions is
\[ 1 \]
Step 4: Final conclusion.
Hence,
\[ \boxed{1} \]
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