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Question

If z=32+i25+32-i25, then


A

Rez=0

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B

Imz=0

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C

Rez>0,Imz>0

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D

Rez>0,Imz<0

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Solution

The correct option is B

Imz=0


Explanation for the correct option:

Find the real and imaginary part of complex number z.

Using cosπ6=32 and sinπ6=12, the right hand side of the equation z=32+i25+32-i25 can be written as:

z=32+i25+32-i25z=cosπ6+isinπ65+cosπ6-isinπ65z=cosπ6+isinπ65+cos-π6+isin-π65[cos-θ=cosθ;sin(-θ)=-sinθ]z=eiπ65+ei-π65[eiθ=cosθ+isinθ]z=ei5π6+ei-5π6

Now, using the formula eiθ+e-iθ=2cosθ the equation can be simplified as

z=ei5π6+ei-5π6z=2cos5π6z=2×-32z=-3 [cos(5π6)=-32]

So the real part of z is Rez=-3<0 and it has no imaginary part. So Imz=0.

Hence, the correct option is B.


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