Question

if z(2-i) = 3+i, find z²⁰.
a. 2¹⁰ b. -2¹⁰
c. 2²⁰ d. -2²⁰
Hope a correct and quick answer from all u geniuses...​

Answer 1

Given :-

• z(2-i) = 3 + i

To Find :-

• $\rm{ {z}^{20} }$

Solution :-

$\rm{\implies z (2-i) = 3+i }$

$\rm{\implies z = \dfrac{ 3+i }{2-i}}$

$\rm{\implies z = \dfrac{ 3+i }{2-i} \times \dfrac{2+i}{2+i}}$

$\rm{\implies z = \dfrac{6+5i + {i}^{2}}{ 4- {i}^{2}}}$

☆ i² = - 1 .

$\rm{\implies z = \dfrac{6+5i -1}{ 4.+1 }}$

$\rm{\implies z = \dfrac{5+5i }{ 5}}$

• $\rm{\implies z = 1+ i }$

In polar form :-

$\rm{\implies z = \sqrt{2} [\dfrac{1}{\sqrt{2} } + \dfrac{1}{\sqrt{2}}i ] }$

$\rm{\implies z = \sqrt{2} [ \cos \dfrac{\pi }{4 } +i \sin \dfrac{\pi}{4} ] }$

• $\rm{ \implies z \: = \sqrt{2} {e}^{i \frac{\pi}{4} } }$

$\rm{ {z }^{20}\: = (\sqrt{2} {e}^{i \frac{\pi}{4} } } {)}^{20}$

$\rm{ \implies {z}^{20} = {2}^{10} . {e}^{i \frac{\pi}{4} \times 20 } }$

$\rm{ \implies {z}^{20} = {2}^{10} . {e}^{i. 5 \pi} }$

$\rm{\implies{ z}^{20} = {2}^{10} [ \cos 5 \pi +i \sin 5 \pi] }$

☆ Cos 5π = -1 and sin 5π = 0

$\rm{\implies {z}^{20} = {2}^{10} [ -1 +0.i ] }$

• $\rm{\implies {z}^{20} = - {2}^{10} }$

Option B .