If z1-2- i and z2-1- i- then -z1-z2-1-z1-z2-i- is

Given: nbsp;z_1=(2 i) and z_2=(1+i) |z_1+z_2+1z_1 z_2+i| = | (2 i)+(1+i)+1(2 i) (1+i)+i| ⇒|z_1+z_2+1z_1 z_2+i|=|4(1 i)| nbsp; ⇒|z_1+z_2+1z_1 z_2+i| =|4 ...

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Byju's Answer

Standard XII

Mathematics

Modulus of a Complex Number

If z1=2-. i a...

Question

If $z_{1} = (2 - i)$ and $z_{2} = (1 + i),$ then $∣ ∣ ∣ \frac{z_{1} + z_{2} + 1}{z_{1} - z_{2} + i} ∣ ∣ ∣$ is

2 \sqrt{2}

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4 \sqrt{2}

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\sqrt{2}

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8 \sqrt{2}

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Solution

The correct option is A $2 \sqrt{2}$
$\frac{z_{1} + z_{2} + 1}{z_{1} - z_{2} + i} = \frac{(2 - i) + (1 + i) + 1}{(2 - i) - (1 + i) + i} = \frac{4}{(1 - i)}$

Now, $∣ ∣ ∣ \frac{z_{1} + z_{2} + 1}{z_{1} - z_{2} + i} ∣ ∣ ∣ = \frac{| 4 |}{| 1 - i |} = 2 \sqrt{2}$

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Similar questions

Q. If

z_{1} = (2 - i)

and

z_{2} = (1 + i),

then

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

is equal to

Q. If

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

, find

∣ ∣ ∣ \frac{z_{1} + z_{2} + 1}{z_{1} + z_{2} + i} ∣ ∣ ∣

If $z_{1} = 2 - i$ and $z_{2} = 1 + i$ , find $∣ ∣ \frac{z_{1} + z_{2} + 1}{z_{1} - z_{2} + 1} ∣ ∣$

Q. If

z_{1} \neq - z_{2}

and

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

then

Q. If

| z_{1} | = | z_{2} | = | z_{3} | = 1

and

z_{1} + z_{2} + z_{3} + = \sqrt{2} + i

, then the complex number

z_{2}_{3} + z_{3} +_{1} + z_{1} +_{2}

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Modulus of a Complex Number

Standard XII Mathematics

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If z1=(2−i) and z2=(1+i), then ∣∣∣z1+z2+1z1−z2+i∣∣∣ is

The correct option is A 2√2 z1+z2+1z1−z2+i=(2−i)+(1+i)+1(2−i)−(1+i)+i=4(1−i) Now, ∣∣∣z1+z2+1z1−z2+i∣∣∣=|4||1−i|=2√2

If $z_{1} = (2 - i)$ and $z_{2} = (1 + i),$ then $∣ ∣ ∣ \frac{z_{1} + z_{2} + 1}{z_{1} - z_{2} + i} ∣ ∣ ∣$ is

The correct option is A $2 \sqrt{2}$
$\frac{z_{1} + z_{2} + 1}{z_{1} - z_{2} + i} = \frac{(2 - i) + (1 + i) + 1}{(2 - i) - (1 + i) + i} = \frac{4}{(1 - i)}$

Now, $∣ ∣ ∣ \frac{z_{1} + z_{2} + 1}{z_{1} - z_{2} + i} ∣ ∣ ∣ = \frac{| 4 |}{| 1 - i |} = 2 \sqrt{2}$