Question

The number of non-zero integral solutions of the equation \[{{\left| 1-i \right|}^{x}}={{2}^{x}}\] is
1) Infinite
2) \[1\]
3) \[2\]
4) None of these

Answer 1

Hint: In this type of question we have to use the concept of complex numbers. We know that the general representation of complex numbers is given by \[z=x+iy\] where \[x\] is called the real part and \[y\] is known as the imaginary part of the complex number \[z\]. The value of \[i\] is defined to be \[i=\sqrt{-1}\]. Also we know that the modulus of a complex number is defined as the square root of the sum of real part’s square and imaginary part’s square i.e. \[\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}\].


Complete step-by-step solution:
Now we have to find the number of non-zero integral solutions of \[{{\left| 1-i \right|}^{x}}={{2}^{x}}\]
Let us consider
\[\Rightarrow {{\left| 1-i \right|}^{x}}={{2}^{x}}\]
Now we know that if \[z=x+iy\] then we can define its modulus as \[\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}\]
\[\begin{align}
  & \Rightarrow {{\left[ \sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}} \right]}^{x}}={{2}^{x}} \\
 & \Rightarrow {{\left( \sqrt{2} \right)}^{x}}={{2}^{x}} \\
\end{align}\]
We will express \[\sqrt{2}\] as \[{{2}^{\dfrac{1}{2}}}\], so that we can write
\[\Rightarrow {{\left( {{2}^{\dfrac{1}{2}}} \right)}^{x}}={{2}^{x}}\]
Now by using the rule of indices \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\] we get,
\[\Rightarrow {{2}^{\dfrac{x}{2}}}={{2}^{x}}\]
As we know that if the bases of both the sides are same then we can compare their indices
\[\begin{align}
  & \Rightarrow \dfrac{x}{2}=x \\
 & \Rightarrow x-\dfrac{x}{2}=0 \\
 & \Rightarrow \dfrac{x}{2}=0 \\
 & \Rightarrow x=0 \\
\end{align}\]
Hence, the equation \[{{\left| 1-i \right|}^{x}}={{2}^{x}}\] does not have a non-zero integral solution.
Thus the number of non-zero integral solutions of the equation \[{{\left| 1-i \right|}^{x}}={{2}^{x}}\] is \[0\]
Therefore, option (4) is the correct option.

Note:In this question students have to remember the definition of modulus of a complex number. Students have to note some of the rules of indices such as \[\sqrt{a}={{a}^{\dfrac{1}{2}}}\], \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\]. Also students have to note that if the bases of both the sides of an equation are the same then they can compare the indices of both the sides.