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

Answer

b
(b) If $(1 - i)^n = 2^n$ ......$(i)$
We know that if two complex numbers are equal, their moduli must also be equal, therefore from $(i)$, we have
$|(1 - i)^n|\, = \,|2^n|$

$ \Rightarrow $ $|1 - i|^n = \,|2|^n$,$(\because \,\,2^n > 0)$
==> $\left[ \sqrt {{1^2} + {{( - 1)}^2}}  \right]^n = 2^n$

==> $(\sqrt 2 )^n = 2^n$
==> $2^{n/2} = 2^n$

==> $\frac{n}{2} = n$

==>$n = 0$
Trick : By inspection, ${(1 - i)^0} = {2^0}\,\,\,\, \Rightarrow 1 = 1$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

If $< {a_n} >$ is an $A.P$. and $a_1 + a_4 + a_7 + .......+ a_{16} = 147$, then $a_1 + a_6 + a_{11} + a_{16}$ is equal to
Let ${\left( { - \,2\, - \,\frac{1}{3}\,i} \right)^3} = \frac{{x \,+ \,iy}}{{27}}(i\, = \,\sqrt { - 1} ),$ where $x$ and $y$ are real numbers, then $y -x$ equals
Any circle through the point of intersection of the lines $x + \sqrt 3 y = 1$ and $\sqrt 3 x - y = 2 $ if intersects these lines at points $P$ and $Q$, then the angle subtended by the arc $PQ$ at its centre is ............ $^o$
Let $S$ be the infinite sum given by $S=\sum \limits_{n=0}^{\infty} \frac{a_n}{10^{2 n}}$ where $\left(a_n\right)_{n \geq 0}$ is a sequence defined by $a_0=a_1=1$ and $a_j=20 a_{j-1}-108 a_{j-2}$ for $j \geq 2$. If $S$ is expressed in the form $\frac{a}{b}$, where $a, b$ are coprime positive integers, then $a$ equals
$\int_{\,0}^{\,\pi /2} {\sin 2x\log \tan x\,dx} $ is equal to
A man tosses a coin $10$ times, scoring $1$ point for each head and $2$ points for each tail. Let $P(K)$ be the probability of scoring at least $K$ points. The largest value of $K$ such that $P(K) > \frac{1}{2}$ is
Let $\mathrm{A}=\{\mathrm{n} \in[100,700] \cap \mathrm{N}: \mathrm{n}$ is neither a multiple of $3$ nor a multiple of 4$\}$. Then the number of elements in $\mathrm{A}$ is
In how many ways can five examination papers be arranged so that physics and chemistry papers never come together
If ${x_n} > {x_{n - 1}} > ... > {x_2} > {x_1} > 1$ then the value of ${\log _{{x_1}}}{\log _{{x_2}}}{\log _{{x_3}}}.....{\log _{{x_n}}}{x_n}^{x_{n - 1}^{{ {\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu \raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1mu}} ^{{x_1}}}}}$ is equal to
If $f(x) = \left\{ \begin{array}{l}\,\,\,\,\,\,\,\,\,\frac{{1 - \cos 4x}}{{{x^2}}},\;\;{\rm{when}}\,x < 0\\\,\,\,\,\,\,\,\,\,\,\,a,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{when}}\,\,x = 0\\\frac{{\sqrt x }}{{\sqrt {(16 + \sqrt x )} - 4}},\,\,{\rm{when}}\,\, x > 0\end{array} \right.$,  is continuous at $x = 0$, then the value of $'a\ '$ will be