MCQ
The value of $\left(\frac{-1+i \sqrt{3}}{1-i}\right)^{30}$ is
  • A
    $2^{15} i$
  • B
    $-2^{15}$
  • $-2^{15} i$
  • D
    $6^{5}$

Answer

Correct option: C.
$-2^{15} i$
c
$\left(\frac{-1+i \sqrt{3}}{1-i}\right)^{30}=\left(\frac{2 \omega}{1-i}\right)^{30}$

$=\frac{2^{30} \cdot \omega^{30}}{\left((1- i )^{2}\right)^{30}}$

$=\frac{2^{30} \cdot 1}{\left(1+ i ^{2}-2 i \right)^{15}}$

$=\frac{2^{30}}{-2^{15} \cdot i ^{15}}$

$=-2^{15} i$

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

Let $f(x) = \int {} {e^x}(x - 1)(x - 2)dx$. Then $f$ decreases in the interval
If $S_1$ and $S_2$ are respectively the sets of local minimum and local maximum points of the function. $f(x) = 9{x^4} + 12{x^3} - 36{x^2} + 25,x \in R$, then
The range of $f (x)$ = $\cos \left[ x \right], - \frac{\pi }{4} < x < \frac{\pi }{4}$ , (where $[.]$ represent greatest integer function less than or equal to $x$ ) is
The temperature $\mathrm{T}(\mathrm{t})$ of a body at time $\mathrm{t}=0$ is $160^{\circ}$ $\mathrm{F}$ and it decreases continuously as per the differential equation $\frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{K}(\mathrm{T}-80)$, where $\mathrm{K}$ is positive constant. If $\mathrm{T}(15)=120^{\circ} \mathrm{F}$, then $\mathrm{T}(45)$ is equal to . . .. . . . . 
If ${x^m}{y^n} = {(x + y)^{m + n}}$ then ${\left. {{{dy} \over {dx}}} \right|_{x = 1,y = 2}}$ is equal to
${d \over {dx}}(\log \tan x) = $
Let $\vec u = a\hat i + b\hat j + c\hat k$  , $\vec v = b\hat i + c\hat j + a\hat k\,\,$  $\vec w = c\hat i + a\hat j + b\hat k = \lambda \vec x + \mu \vec y$ where $\left[ {\vec u\,\,\vec v\,\,\vec w} \right] = 0\,\  \,\,\left( {a + b + c} \right),\,\,\lambda ,\mu  \ne 0$ then the vectors $\vec x,\vec y,\vec u,\vec v,\vec w$ are
If ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y + {\tan ^{ - 1}}z = \pi $, then $x + y + z$ is equal to
Period of $\cot 3x - \cos (4x + 3)$ is
The value of $\int_0^{{{\sin }^2}x} {{{\sin }^{ - 1}}\sqrt t \,dt + \int_0^{{{\cos }^2}x} {{{\cos }^{ - 1}}\sqrt t \,dt} } $ is