Lf z1 and z2 are purely real then z1- z 2 - z1z2 form a

The correct option is D straight lineIf z_1 and z_2 are purely real.Then z_1=z_1 and z_2=z_2 Thus the points z_1,z_2,z_1,z_2 will b ...

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Standard IX

Mathematics

Condition That 2 Lines Are Coplanar

lf z1 and ...

Question

lf $z_{1}$ and $z_{2}$ are purely real then $z_{1},_{2},_{1} z_{2}$ form a

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straight line

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Solution

The correct option is D straight line
If $z_{1}$ and $z_{2}$ are purely real.
Then
$z_{1} =_{1}$ and
$z_{2} =_{2}$
Thus the points
$z_{1},_{2},_{1}, z_{2}$ will be collinear, thus forming a straight line.

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Similar questions

Q. If

z_{1}

and

z_{2}

be complex numbers such that

z_{1} + i (_{2}) = 0

and

a r g (_{1} z_{2}) = \frac{π}{3}

. Then,

a r g (_{1})

is equal to

Q. Let

z_{1}

and

z_{2}

be two complex numbers such that

\frac{z_{1} - 3 z_{2}}{3 - z_{1} {¯ ¯ ¯ z}_{2}}

is unimodular. If

z_{2}

is not unimodular then

| z_{1} |

Q. I.

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

\frac{z_{1}}{z_{2}}

is purely imaginary
II. If z is purely real then

z = ¯ ¯ ¯ z

Q. For any two complex numbers

z_{1}

and

z_{2}

with

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

{∣ ∣ \sqrt{2} z_{1} + i \sqrt{3}_{2} ∣ ∣}^{2} + {∣ ∣ \sqrt{3}_{1} + i \sqrt{2} z_{2} ∣ ∣}^{2}

For any complex numbers $z, z_{1}$ and $z_{2}$ prove that:
(i) $a r g (¯ z) = - a r g (z)$
(ii) $a r g (z_{1} z_{2}) = a r g (z_{1}) + a r g (z_{2})$
(iii) $a r g (z_{1}_{2}) = a r g (z_{1}) - a r g (z_{2})$
(iv) $a r g (\frac{z_{1}}{z_{2}}) = a r g (z_{1}) - a r g (z_{2})$

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lf z1 and z2 are purely real then z1, ¯¯¯¯¯z2,¯¯¯¯¯z1z2 form a

The correct option is D straight lineIf z1 and z2 are purely real.Thenz1=¯¯¯¯¯z1 and z2=¯¯¯¯¯z2Thus the points z1,¯¯¯¯¯z2,¯¯¯¯¯z1,z2 will be collinear, thus forming a straight line.

lf $z_{1}$ and $z_{2}$ are purely real then $z_{1},_{2},_{1} z_{2}$ form a

The correct option is D straight line
If $z_{1}$ and $z_{2}$ are purely real.
Then
$z_{1} =_{1}$ and
$z_{2} =_{2}$
Thus the points
$z_{1},_{2},_{1}, z_{2}$ will be collinear, thus forming a straight line.