If z 1-2- i and z 2-1- i- find -1-1-z2-1-z1-z2-1-

Here z_1=2 i and z_2=1+i ∴ | z_1+z_2+1/z_1 z_2+1 | = | 2 i+1+i+1/2 i 1 i+1 |= | 4/2 2i | = | 4/2 2i |=4/√((2)^2+( 2)^2) =4/√(4+4)=4/√(8)=4/2√(2)=√(2) ...

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Byju's Answer

Standard XI

Mathematics

Inequality of 2 Complex Numbers

If z 1=2- i a...

Question

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

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Solution

Here $z_{1} = 2 - i$ and $z_{2} = 1 + i$

$∴ ∣ ∣ \frac{z_{1} + z_{2} + 1}{z_{1} - z_{2} + 1} ∣ ∣$

$= ∣ ∣ \frac{2 - i + 1 + i + 1}{2 - i - 1 - i + 1} ∣ ∣ = ∣ ∣ \frac{4}{2 - 2 i} ∣ ∣$

$= ∣ ∣ \frac{4}{2 - 2 i} ∣ ∣ = \frac{4}{\sqrt{(2)^{2} + (- 2)^{2}}}$

$= \frac{4}{\sqrt{4 + 4}} = \frac{4}{\sqrt{8}} = \frac{4}{2 \sqrt{2}} = \sqrt{2}$

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If z1=2−i and z2=1+i, find ∣∣z1+z2+1z1−z2+1∣∣

Here z1=2−i and z2=1+i ∴ ∣∣z1+z2+1z1−z2+1∣∣ =∣∣2−i+1+i+12−i−1−i+1∣∣=∣∣42−2i∣∣ =∣∣42−2i∣∣=4√(2)2+(−2)2 =4√4+4=4√8=42√2=√2

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