If z_{1} = 2 - i and z_{2} = 1 + i,find left| | vedclass

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(B) Given $z_{1} = 2 - i$ and $z_{2} = 1 + i$. First,calculate the numerator: $z_{1} + z_{2} + 1 = (2 - i) + (1 + i) + 1 = 4$. Next,calculate the deno

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$\sqrt{2}$

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(B) Given $z_{1} = 2 - i$ and $z_{2} = 1 + i$. First,calculate the numerator: $z_{1} + z_{2} + 1 = (2 - i) + (1 + i) + 1 = 4$. Next,calculate the denominator: $z_{1} - z_{2} + 1 = (2 - i) - (1 + i) + 1 = 2 - i - 1 - i + 1 = 2 - 2i$. Now,substitute these into the expression: $\left| \frac{4}{2 - 2i} \right| = \left| \frac{4}{2(1 - i)} \right| = \left| \frac{2}{1 - i} \right|$. To simplify,multiply the numerator and denominator by the conjugate $(1 + i)$: $\left| \frac{2(1 + i)}{(1 - i)(1 + i)} \right| = \left| \frac{2(1 + i)}{1^2 - i^2} \right|$. Since $i^2 = -1$,we have $\left| \frac{2(1 + i)}{1 + 1} \right| = \left| \frac{2(1 + i)}{2} \right| = |1 + i|$. The modulus is $|1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}$.

If $z_{1} = 2 - i$ and $z_{2} = 1 + i$,find $\left| \frac{z_{1} + z_{2} + 1}{z_{1} - z_{2} + 1} \right|$.

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