IIT JEE 1982 Maths - MCQ Question 7

Algebra Level 3

z = ( 3 2 + i 2 ) 5 + ( 3 2 i 2 ) 5 z = \left(\dfrac{\sqrt3}2+\dfrac{i}2\right)^5 + \left(\dfrac{\sqrt3}2-\dfrac{i}2\right)^5

For z z as defined above, which one of the options is correct?

Notation: i = 1 i=\sqrt{-1} is the imaginary unit .


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Re(z)>0, Im(z)<0 Re(z)>0, Im(z)>0 Im(z)=0 Re(z)=0

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1 solution

z = ( c o s ( π 6 ) + i s i n ( π 6 ) ) 5 + ( c o s ( π 6 ) + i s i n ( π 6 ) ) 5 z= ( cos ( \frac {\pi} {6}) +i sin ( \frac {\pi} {6}))^5+( cos (- \frac {\pi} {6}) +i sin (- \frac {\pi} {6}))^5

z = ( c o s ( 5 π 6 ) + i s i n ( 5 π 6 ) ) + ( c o s ( 5 π 6 ) + i s i n ( 5 π 6 ) ) z= ( cos ( \frac {5\pi} {6}) +i sin ( \frac {5\pi} {6}))+( cos (- \frac {5\pi} {6}) +i sin (- \frac {5\pi} {6})) ....De Moivre's Theorem

z = 2 c o s ( 5 π 6 ) z=2 cos ( \frac {5\pi} {6})

so z is real.

A solution through Binomial theorem is also possible as all the complex terms will have opposite signs in the expansion. Another solution would be to use ω \omega and its properties as the above numbers are i ω -i\omega and i ω i\omega respectively.

Ajinkya Shivashankar - 4 years, 6 months ago

Can you give me the link where i can learn this stuff ? (i mean brilliant wiki link)

Daniel Sugihantoro - 4 years, 6 months ago

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Check this out

Steven Chase - 4 years, 6 months ago

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ah thank you so much

Daniel Sugihantoro - 4 years, 6 months ago

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