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Please,please help me - Complex numbers and quadratic equations - JEE Main-6

Let z= \left ( a^{2}-3a+2 \right )+i\left ( a^{2}-1 \right ) is a purely imaginary complex number, with Im\left ( z \right )\neq 0 then 'a' equals \left ( Where\; a\: \epsilon \: R \right )

  • Option 1)

    1

  • Option 2)

    2

  • Option 3)

    3

  • Option 4)

    4

 
Answers (1)
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 Since z is purely imaginary, So Re(z)=0

\Rightarrow a^{2}-3a+2=0\: \Rightarrow \, a=1, \: or\: a=2

But it is given Im\left ( z \right )\neq 0 \: so\: a\neq1

Hence a=2

\therefore Option (B)

 

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

 

 

 


Option 1)

1

This is incorrect

Option 2)

2

This is correct

Option 3)

3

This is incorrect

Option 4)

4

This is incorrect

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