CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Addition of Vectors

If z1 & z2 ...

Question

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, $z_{1}, z_{2}, z_{1} + z_{2}$ is

A

a parallelogram but not a rectangle or a rhombus

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

B

a rectangle but not a square

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

C

a rhombus but not a square

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

D

a square

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

Open in App

Solution

The correct option is C a rhombus but not a square
$a r g (\frac{z_{1} + z_{2}}{z_{1} - z_{2}}) = \frac{π}{2}$
$| z_{1} + z_{2} | \neq | z_{1} - z_{2} |$
$∴$ Figure formed by $0, z_{1} z_{2}, z_{1} + z_{2}$
$\Rightarrow a r g (z_{1} + z_{2}) - a r g (z_{1} - z_{2}) = \frac{π}{2}$
$∴ a r g (z_{1} + z_{2}) = \frac{π}{2} + a r g (z_{1} - z_{2})$
$\Rightarrow z_{1} = r_{1} e^{i Q_{1}}$ $z_{2} = r_{2} e^{i Q_{2}}$ $z_{3} = r_{3} e^{i Q_{3}}$ $z_{4} = r_{4} e^{i Q_{4}}$
$\Rightarrow z_{1} + z_{2} = r_{1} e^{i Q_{1}} + r_{2} e^{i Q_{2}}$
$\Rightarrow z_{1} - z_{2} = r_{1} e^{i Q_{1}} - r_{2} e^{i Q_{2}}$
$∴$ The points represent a rhombus.
Hence, the answer is a rhombus but not a square.

Suggest Corrections

thumbs-up

0

Similar questions

z_{1}

z_{2}

are two complex numbers, then prove that

| z_{1} | + | z_{2} | = ∣ ∣ ∣ \frac{z_{1} + z_{2}}{2} + \sqrt{z_{1} z_{2}} ∣ ∣ ∣ + ∣ ∣ ∣ \frac{z_{1} + z_{2}}{2} - \sqrt{z_{1} z_{2}} ∣ ∣ ∣

z_{1}

z_{2}

are two complex numbers such that

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

a r g z_{1} + a r g z_{2} = π

z_{1}

z_{2}

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}}

z_{1}, z_{2}

are the complex numbers such that

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

a r g z_{1} - a r g z_{2}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Vector Addition

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Addition of Vectors

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon