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Let $z= \left ( a-2 \right )+i\left ( 5-a \right )$ , p is $Re \: \left ( z \right )$ when z is purely real and q is $Im\left ( z \right )$ when z is purely imaginary, then $p+q$ equals $\left ( Where\: a\, \epsilon \, R \right )$

Option 1)
$3$

Option 2)
$4$

Option 3)
$5$

Option 4)
$6$

Answers (1)

best_answer

When z is purely real then $5-a=0$

$\Rightarrow \: a=5$

At that time $p=Re\left ( z \right )=\left ( a-2 \right )=5-2=3$

When z is purely imaginary then $a-2=0$

$\Rightarrow \: a=2$

At that time $q=Im\left ( z \right )=5-a=5-2=3$

$\therefore \: p+q=3+3=6$

$\therefore$ Option (D)

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)

$3$

This is incorrect

Option 2)

$4$

This is incorrect

Option 3)

$5$

This is incorrect

Option 4)

$6$

This is correct

Posted by

divya.saini

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