If z - x - iy then z - x 2- y 2 z 1-2- i - z 2-1- i- find - z 1- z 2-1- z 1- z 2- i - -

Ifz + x + iy then (z) = √(x^2+y^2) We have, z_1=2 i,z_2=1+i z_1+z_2=2 i+1+i=3 And z_1 z_2=2 i 1 i=1 2i z_1+z_2+1/z_1 z_2+i=3+1/1 2i+i =4/1 i =4/1 i×1+i/1+i ...

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Byju's Answer

Standard XII

Mathematics

Modulus of a Complex Number

If z + x + iy...

Question

$If z + x + i y then (z) = \sqrt{x^{2} + y^{2}} z_{1} = 2 - i, z_{2} = 1 + i, find ∣ ∣ \frac{z_{1} + z_{2} + 1}{z_{1} - z_{2} + i} ∣ ∣ .$

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Solution

$If z + x + i y then (z) = \sqrt{x^{2} + y^{2}} We have, z_{1} = 2 - i, z_{2} = 1 + i z_{1} + z_{2} = 2 - i + 1 + i = 3 A n d z_{1} - z_{2} = 2 - i - 1 - i = 1 - 2 i \frac{z_{1} + z_{2} + 1}{z_{1} - z_{2} + i} = \frac{3 + 1}{1 - 2 i + i} = \frac{4}{1 - i} = \frac{4}{1 - i} \times \frac{1 + i}{1 + i} = \frac{4 (1 + i)}{1^{2} + 1^{2}} = \frac{4 (1 + i)}{2} = 2 (1 + i) ∴ ∣ ∣ \frac{z_{1} + z_{2} + 1}{z_{1} - z_{2} + i} ∣ ∣ = | 2 (1 + i) | = | 2 | | 1 + i | (∵ | z_{1} z_{2} | = | z_{1} | \times | z_{2} |) = 2 \times \sqrt{1^{2} + 1^{2}} = 2 \times \sqrt{2} = 2 \sqrt{2}$

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Similar questions

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

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. If

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

, find

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

Q. If

z_{1} = (2 - i)

and

z_{2} = (1 + i),

then

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

Q. If

| z | = 1

and

\frac{z_{1} - z_{3}}{z_{2} - z_{3}} = \frac{z - i}{z + i}

then

z_{1}, z_{2}, z_{3}

will be vertices of a

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If z+x+iy then (z)=√x2+y2 z1=2−i,z2=1+i, find ∣∣z1+z2+1z1−z2+i∣∣.

Ifz+x+iy then (z)=√x2+y2We have,z1=2−i,z2=1+iz1+z2=2−i+1+i=3And z1−z2=2−i−1−i=1−2iz1+z2+1z1−z2+i=3+11−2i+i=41−i=41−i×1+i1+i=4(1+i)12+12=4(1+i)2=2(1+i)∴ ∣∣z1+z2+1z1−z2+i∣∣=|2(1+i)|=|2||1+i| (∵ |z1z2|=|z1|×|z2|)=2×√12+12=2×√2=2√2

$If z + x + i y then (z) = \sqrt{x^{2} + y^{2}} z_{1} = 2 - i, z_{2} = 1 + i, find ∣ ∣ \frac{z_{1} + z_{2} + 1}{z_{1} - z_{2} + i} ∣ ∣ .$