If (1 - i)^n = 2^n,then n = | vedclass

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(B) Given the equation $(1 - i)^n = 2^n$ $(i)$.Taking the modulus on both sides,we have $|(1 - i)^n| = |2^n|$.Since $|z^n| = |z|^n$,we get $|1 - i|^n

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(B) Given the equation $(1 - i)^n = 2^n$ $(i)$.Taking the modulus on both sides,we have $|(1 - i)^n| = |2^n|$.Since $|z^n| = |z|^n$,we get $|1 - i|^n = 2^n$ (as $2^n > 0$ for real $n$).The modulus of $1 - i$ is $\sqrt{1^2 + (-1)^2} = \sqrt{2}$.Substituting this,we get $(\sqrt{2})^n = 2^n$.This simplifies to $(2^{1/2})^n = 2^n$,which means $2^{n/2} = 2^n$.Equating the exponents,we get $\frac{n}{2} = n$,which implies $n - \frac{n}{2} = 0$,so $\frac{n}{2} = 0$,hence $n = 0$.Verification: $(1 - i)^0 = 1$ and $2^0 = 1$. Since $1 = 1$,$n = 0$ is the correct solution.

If $(1 - i)^n = 2^n$,then $n = $

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