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How do you use DeMoivre’s theorem to simplify ${( - 1 + i)^{10}}$?

Last updated date: 10th Aug 2024
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Hint: Try to convert the complex number into polar form first. Then apply the DeMoivre’s theorem and write ${(i - 1)^{10}}$ in the complex form of $a + bi$, where $a\,\,and\,\,b$ are the real numbers and these numbers do not use the trigonometric function. After rearranging the term and getting in the general form for the theorem we can proceed with the question.

Complete step by step solution:
The given question is ${( - 1 + i)^{10}}$. ${( - 1 + i)^{10}}$ can be also written as ${(i - 1)^{10}}$. First write the complex number in polar form, that is in the form of $a + bi$ and then apply the De-Moivre’s theorem .De-Moivre’s theorem states that:
$\Rightarrow x + iy = \sqrt {{x^2} + {y^2}{e^{i\theta }}}$ $where\,{e^{i\theta }} = cos(\phi ) + i\sin (\phi )$;$\phi = \arctan \left( {\dfrac{y}{x}} \right)$
Applying this to our question and after solving, we get:
$\Rightarrow - 1 + i = \sqrt {2{e^{ - 1\dfrac{\pi }{4}}}}$
Now putting the value in our question, proceeding further, on solving we get:
$\Rightarrow {(i - 1)^{10}} = {(\sqrt {2{e^{ - i\dfrac{\pi }{4}}}} )^{10}} = {(\surd 2)^{10}}{e^{ - i\dfrac{{10\pi }}{4}}}$
But,${e^{ - i\dfrac{{10\pi }}{4}}} = {e^{ - i\dfrac{{8\pi }}{4}}} \cdot {e^{ - i\dfrac{\pi }{2}}} = {e^{ - i\dfrac{\pi }{4}}}$
$\Rightarrow {(\sqrt 2 )^{10}} = 32$
$\therefore {(i - 1)^{10}} = - 32i$
So, this is the final answer in the polar form of the given complex number.

Hence, we finally get ${(i - 1)^{10}} = - 32i$ as the answer.