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Standard XII

Mathematics

Complex Numbers

The complex n...

Question

The complex numbers $z_{1} = 8 + 9 i, z_{2} = 4 - 6 i$ then $z_{1} - z_{2}$

4 + 15 i

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4 - 3 i

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12 + 3 i

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12 - 15 i

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Solution

The correct option is A $4 + 15 i$
$z_{1} = 8 + 9 i z_{2} = 4 - 6 i z_{1} - z_{2} = 8 + 9 i - 4 + 6 i = 4 + 15 i$

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Similar questions

Q. Let

z_{1} = 10 + 6 i

and

z_{2} = 4 + 6 i

. If

z

is any complex number such that the argument of

(z - z_{1}) / (z - z_{2})

π / 4

, then

| z - 7 - 9 i |

z_{1} = 4 - 3 i, z_{2} = - 1 + 2 i

then

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

Q. Let

z_{1} = 10 + 6 i

and

z_{2} = 4 + 6 i,

where

i = \sqrt{- 1} .

z

is any complex number such that

arg (\frac{z - z_{1}}{z - z_{2}}) = \frac{π}{4},

then the value of

| z - 7 - 9 i |

Q. Let

z_{1} = 10 + 6 i

and

z_{2} = 4 + 6 i

If z is any complex number such that the argument of

(\frac{z - z_{1}}{z - z_{2}})

\frac{π}{4}

then prove that

| z - 7 - 9 i | = 3 \sqrt{2}

Let $z_{1} = 10 + 6 i$ and $z_{2} = 4 + 6 i$ . If $z$ is a complex number such that the argument of $(z - z_{1}) / (z - z_{2})$ is $π / 4$ , then prove that $| z - 7 - 9 i | = 3 \sqrt{2}$ .

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The complex numbers z1=8+9i,z2=4−6i then z1−z2

The correct option is A 4+15iz1=8+9iz2=4−6iz1−z2=8+9i−4+6i=4+15i

The complex numbers $z_{1} = 8 + 9 i, z_{2} = 4 - 6 i$ then $z_{1} - z_{2}$

The correct option is A $4 + 15 i$
$z_{1} = 8 + 9 i z_{2} = 4 - 6 i z_{1} - z_{2} = 8 + 9 i - 4 + 6 i = 4 + 15 i$