If z 1-2- i and z 2-1- i - find z1-22-1-1-2-z-2- i -

Z_1+z_2+1/z_1 z_2+i ⇒(2 i )+(1+i )+1/(2 i ) (1+i )+i ⇒4/(1+i )×(1 i )/(1 i ) = 2(1 i) ⇒ | z_1+z_2+1/z_1 z_2+i | = 2 ×√(1+ ( 1)^2 ) = 2 √(2) ...

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Inequality of 2 Complex Numbers

If z 1=2- i a...

Question

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

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Solution

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z_{1} = (2 - i)

z_{2} = (1 + i),

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

z_{1} = (2 - i)

z_{2} = (1 + i),

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

z_{1} = 1 - 3 i

z_{2} = 2 + i

{¯ z}_{1} + {¯ z}_{2} =

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

z_{1} = \frac{1}{a + i}, a \neq 0

z_{2} = \frac{1}{1 + b i}, b \neq 0

z_{1} = {¯ z}_{2}

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