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

If z(2−i)=3+iz(2 - i) = 3 + i then z20={z^{20}} = :
A. −1024 - 1024
B. 1−i1 - i
C. 1+i1 + i
D. 10241024

Explanation

Solution

First try to find out the zz after getting rationalise it and try to convert it as z=∣z∣(cos⁡θ+isin⁡θ)z = \left| z \right|\left( {\cos \theta + i\sin \theta } \right) because this can be also written as z=∣z∣eiθz = \left| z \right|{e^{i\theta }} Now we have to find out the z20{z^{20}} so take 2020 on both sides and solve according to that .

Complete step-by-step answer:
In this question first try to find out the z for this ,
z(2−i)=3+iz(2 - i) = 3 + i
Now transfer 2−i2 - i to the RHS , or in denominator of RHS ,
we get
⇒\Rightarrow z=3+i2−iz = \dfrac{{3 + i}}{{2 - i}}
Now multiply and divide 2+i2 + i in both numerator and denominator ,
⇒\Rightarrow z=3+i2−i×2+i2+iz = \dfrac{{3 + i}}{{2 - i}} \times \dfrac{{2 + i}}{{2 + i}}
On solving we get ,
⇒\Rightarrow z=6+3i+2i+i222−i2z = \dfrac{{6 + 3i + 2i + {i^2}}}{{{2^2} - {i^2}}}
We know that the value of i2=−1{i^2} = - 1 hence put it on equation
⇒\Rightarrow z=5+5i5z = \dfrac{{5 + 5i}}{5}
or z=1+iz = 1 + i
Now try to convert z=∣z∣(cos⁡θ+isin⁡θ)z = \left| z \right|\left( {\cos \theta + i\sin \theta } \right) because this can be also written as z=∣z∣eiθz = \left| z \right|{e^{i\theta }} so for this take common 2\sqrt 2 from the equation z=1+iz = 1 + i
⇒\Rightarrow z=2(12+12i)z = \sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }}i} \right)
As we know that the cos⁡π4=12=sin⁡π4\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} = \sin \dfrac{\pi }{4} hence
⇒\Rightarrow z=2(cos⁡π4+isin⁡π4)z = \sqrt 2 \left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)
It can also be written as z=2eiπ4z = \sqrt 2 {e^{i\dfrac{\pi }{4}}}
Now we have to find out the z20{z^{20}} so ,
⇒\Rightarrow z20=(2eiπ4)20{z^{20}} = {\left( {\sqrt 2 {e^{i\dfrac{\pi }{4}}}} \right)^{20}}
⇒\Rightarrow z20=((2)20eiπ4×20){z^{20}} = \left( {{{\left( {\sqrt 2 } \right)}^{20}}{e^{i\dfrac{\pi }{4} \times 20}}} \right)
⇒\Rightarrow z20=(210ei5π){z^{20}} = \left( {{2^{10}}{e^{i5\pi }}} \right)
So 210=1024{2^{10}} = 1024 and we can write ei5π=cos⁡5π+isin⁡5π{e^{i5\pi }} = \cos 5\pi + i\sin 5\pi
⇒\Rightarrow z20=1024(cos⁡5π+isin⁡5π){z^{20}} = 1024\left( {\cos 5\pi + i\sin 5\pi } \right)
And as we know that the value of cos⁡5π=−1\cos 5\pi = - 1 and sin⁡5π=0\sin 5\pi = 0 therefore ,
⇒\Rightarrow z20=−1024{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 (cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ{\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 ω\omega is the cube root of unity
 ω3n=(ω3)n=1n=1 ω3n+1=ω3n.ω=1.ω=ω ω3n+2=ω3n.ω2=1.ω2=ω2  \ {\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} \\\ \