Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# If $z(2 - i) = 3 + i$ then ${z^{20}} =$ :A. $- 1024$ B. $1 - i$ C. $1 + i$ D. $1024$

Last updated date: 20th Jun 2024
Total views: 405k
Views today: 7.05k
Verified
405k+ views
Hint: First try to find out the $z$ after getting rationalise it and try to convert it as $z = \left| z \right|\left( {\cos \theta + i\sin \theta } \right)$ because this can be also written as $z = \left| z \right|{e^{i\theta }}$ Now we have to find out the ${z^{20}}$ so take $20$ on both sides and solve according to that .

In this question first try to find out the z for this ,
$z(2 - i) = 3 + i$
Now transfer $2 - i$ to the RHS , or in denominator of RHS ,
we get
$\Rightarrow$$z = \dfrac{{3 + i}}{{2 - i}} Now multiply and divide 2 + i in both numerator and denominator , \Rightarrow$$z = \dfrac{{3 + i}}{{2 - i}} \times \dfrac{{2 + i}}{{2 + i}}$
On solving we get ,
$\Rightarrow$$z = \dfrac{{6 + 3i + 2i + {i^2}}}{{{2^2} - {i^2}}} We know that the value of {i^2} = - 1 hence put it on equation \Rightarrow$$z = \dfrac{{5 + 5i}}{5}$
or $z = 1 + i$
Now try to convert $z = \left| z \right|\left( {\cos \theta + i\sin \theta } \right)$ because this can be also written as $z = \left| z \right|{e^{i\theta }}$ so for this take common $\sqrt 2$ from the equation $z = 1 + i$
$\Rightarrow$$z = \sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }}i} \right) As we know that the \cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} = \sin \dfrac{\pi }{4} hence \Rightarrow$$z = \sqrt 2 \left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)$
It can also be written as $z = \sqrt 2 {e^{i\dfrac{\pi }{4}}}$
Now we have to find out the ${z^{20}}$ so ,
$\Rightarrow$${z^{20}} = {\left( {\sqrt 2 {e^{i\dfrac{\pi }{4}}}} \right)^{20}} \Rightarrow$${z^{20}} = \left( {{{\left( {\sqrt 2 } \right)}^{20}}{e^{i\dfrac{\pi }{4} \times 20}}} \right)$
$\Rightarrow$${z^{20}} = \left( {{2^{10}}{e^{i5\pi }}} \right) So {2^{10}} = 1024 and we can write {e^{i5\pi }} = \cos 5\pi + i\sin 5\pi \Rightarrow$${z^{20}} = 1024\left( {\cos 5\pi + i\sin 5\pi } \right)$
And as we know that the value of $\cos 5\pi = - 1$ and $\sin 5\pi = 0$ therefore ,
$\Rightarrow$${z^{20}} = - 1024$

Hence option A is the correct answer.

Note: De Moivre’s Theorem for integral index state that If n is a integer, then ${\left( {\cos \theta + i\sin \theta } \right)^n} = \cos n\theta + i\sin n\theta$ we will use this proof in solving the question .
In general, if n be a positive integer then, where $\omega$ is the cube root of unity
$\ {\omega ^{3n}} = {({\omega ^3})^n} = {1^n} = 1 \\ {\omega ^{3n + 1}} = {\omega ^{3n}}.\omega = 1.\omega = \omega \\ {\omega ^{3n + 2}} = {\omega ^{3n}}.{\omega ^2} = 1.{\omega ^2} = {\omega ^2} \\ \$