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  • Can someone help me with this, - Complex numbers and quadratic equations - JEE Main-6

Lets z\neq -i  be any complex number such that  \frac{z-i}{z+i}     is a purely imaginary number

Then z+\frac{1}{z}  is:

  • Option 1)

    0

  • Option 2)

     any non-zero real number other than 1.
     

  • Option 3)

     any non-zero real number.

     

  • Option 4)

     a purely imaginary number.

 

Answers (2)

best_answer

As we have learned

Purely Imaginary Complex Number -

z=x+iy, \boldsymbol{x=0}, y\epsilon R

& i2=-1

- wherein

Real part of z = Re (z) = x & Imaginary part of z = Im (z) = y

 

 \frac{z-i}{z+i}= ik , k \neq 0

\Rightarrow z-i = ikz-k

\Rightarrow z(1-ik)= i-k

\Rightarrow z= -k+i/ (1-ik)

\Rightarrow 1/z= (1-ik)/ i-k

So, z+\frac{1}{z}= \frac{(i-k)}{(1-ik)}+ \frac{(1-ik)}{(i-k)}

=\frac{-1 +k^2-2i k +1-k^2-2ik}{i -k +k+ik^2}= \frac{-4ik}{i(1+k^2)}= \frac{-4k}{1+k^2}

 

 

 

 

 

 


Option 1)

0

Option 2)

 any non-zero real number other than 1.
 

Option 3)

 any non-zero real number.

 

Option 4)

 a purely imaginary number.

Posted by

Himanshu

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