Question

\(If \,z_1=2-i,z_2=1+i, find \,|\frac{z_1+z_2+1}{z1-z2+1}|.\)

Answer 1

\(z_1=2-i,z_2=1+i\)

\(|\frac{z_1+z_2+1}{z1-z2+1}=|\frac{(2-i)+(1+i)+1}{(2-i)+(1+i)+1}\)

\(|\frac{4}{2-2i}=|\frac{4}{2(1-i)}\)

\(=|\frac{2}{1-i}×\frac{1+i}{1+i}|=|\frac{2(1+i)}{1^2-i^2}|\)

\(=\frac{2(1+i)}{1+i)}|\)   \([i^2=-1]\)

\(|\frac{2(1+i)}{2}|\)

\(|1+i|=\sqrt1^2+1^2=\sqrt2\)

Thus,the value of \(|\frac{z_1+z_2+1}{z_1-z_2+1}]\,is\,\sqrt2\).