Answer:
The real and imaginary parts of the result are [tex]\frac{1441}{1601}[/tex] and [tex]\frac{4}{1601}[/tex], respectively.
Step-by-step explanation:
Let be [tex]f(z) = \frac{z}{z+1}[/tex], the following expression is expanded by algebraic means:
[tex]f(z) = \frac{z\cdot (z-1)}{(z+1)\cdot (z-1)}[/tex]
[tex]f(z) = \frac{z^{2}-z}{z^{2}-1}[/tex]
[tex]f(z) = \frac{z^{2}}{z^{2}-1}-\frac{z}{z^{2}-1}[/tex]
If [tex]z = 4 - i5[/tex], then:
[tex]z^{2} = (4-i5)\cdot (4-i5)[/tex]
[tex]z^{2} = 16-i20-i20-(-1)\cdot (25)[/tex]
[tex]z^{2} = 41 - i40[/tex]
Then, the variable is substituted in the equation and simplified:
[tex]f(z) = \frac{41-i40}{41-i39} -\frac{4-i5}{41-i39}[/tex]
[tex]f(z) = \frac{37-i35}{41-i39}[/tex]
[tex]f(z) = \frac{(37-i35)\cdot (41+i39)}{(41-i39)\cdot (41+i39)}[/tex]
[tex]f(z) = \frac{1517-i1435+i1443+1365}{3202}[/tex]
[tex]f(z) = \frac{2882+i8}{3202}[/tex]
[tex]f(z) = \frac{1441}{1601} + i\frac{4}{1601}[/tex]
The real and imaginary parts of the result are [tex]\frac{1441}{1601}[/tex] and [tex]\frac{4}{1601}[/tex], respectively.