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Q

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)

$4$

92 Views

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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