If for the complex numbers $z_{1}$ and $z_{2}$ , $| z_{1} + z_{2} | = | z_{1} - z_{2} |$ , then $a m p (z_{1}) \sim a m p (z_{2}) =$

π

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{π}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $\frac{π}{2}$
Let $z_{1} = x_{1} + i y_{1}$ and $z_{2} = x_{2} + i y_{2}$

Now $z_{1} + z_{2} = (x_{1} + x_{2}) + i (y_{1} + y_{2})$

and $z_{1} - z_{2} = (x_{1} - x_{2}) + i (y_{1} - y_{2})$

As $| z_{1} + z_{2} | = | z_{1} - z_{2} |$ , we get

${(x_{1} + x_{2})}^{2} + {(y_{1} + y_{2})}^{2} = {(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}$

$\Rightarrow x_{1} x_{2} + y_{1} y_{2} = 0$ ...(1)

Now, $a m p (z_{1}) - a m p (z_{2}) = {tan}^{- 1} \frac{y_{1}}{x_{1}} - {tan}^{- 1} \frac{y_{2}}{x_{2}}$

$= {tan}^{- 1} ⎛ ⎝ \frac{\frac{y_{1}}{x_{1}} - \frac{y_{2}}{x_{2}}}{1 + \frac{y_{1}}{x_{1}} \frac{y_{2}}{x_{2}}} ⎞ ⎠ = {tan}^{- 1} (\frac{x_{2} y_{1} - y_{2} x_{1}}{x_{1} x_{2} + y_{1} y_{2}}) = {tan}^{- 1} \infty$ [From (i)]

$∴ | a m p (z_{1}) - a m p (z_{2}) | = \frac{π}{2}$

Suggest Corrections

Similar questions

If | $Z_{1}$ + $Z_{2}$ | = | $Z_{1}$ | - | $Z_{2}$ | where $Z_{1}$ , $Z_{2}$ are two non-zero complex numbers, then arg( $Z_{1}$ ) - arg( $Z_{2}$ ) is

Q. If z₁ and z₂ are two complex numbers such that

|z_{1}| = |z_{2}|

and arg(z₁) + arg(z₂) =

π

, then show that

z_{1} = - \bar{z_{2}}

Q. If z₁ and z₂ are two non-zero complex such that

∣ z_{1} + z_{2} ∣=∣ z_{1} ∣ + ∣ z_{2} ∣,

then arg

(z_{1}) -

arg

(z_{2})

is equal to

if $Z_{1}$ and $Z_{2}$ are 2 non - complex number such that | $Z_{1}$ + $Z_{2}$ | = | $Z_{1}$ | + | $Z_{2}$ |,Then arg ( $Z_{1}$ ) - arg( $Z_{2}$ ) is equal to

Hint: Square on both sides.

Q. If

z_{1} & z_{2}

are two complex numbers and

a r g (\frac{z_{1} + z_{2}}{z_{1} - z_{2}}) = \frac{π}{2}

but

| z_{1} + z_{2} | \neq | z_{1} - z_{2} |

, then the figure formed by the points represented by 0,

z_{1}, z_{2}, z_{1} + z_{2}

Join BYJU'S Learning Program

If for the complex numbers z1 and z2, |z1+z2|=|z1−z2| , then amp(z1)∼amp(z2)=

If for the complex numbers $z_{1}$ and $z_{2}$ , $| z_{1} + z_{2} | = | z_{1} - z_{2} |$ , then $a m p (z_{1}) \sim a m p (z_{2}) =$