f(z)=Re(z)+Im(z)
f(z)=x+iy
I got three case for the value of h
When h is real. h=a+0i. a is a real number
$$\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}$$
$$\lim_{h\to 0}\frac{x+a+y-(x+y)}{a}=\frac{a}{a}=1$$
When h is imiginary. h=0+bi. b is a real number $$\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}$$ $$\lim_{h\to 0}\frac{x+y+b-(x+y)}{b}=\frac{b}{b}=1$$ When h=a+bi $$\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}$$ $$\lim_{h\to 0}\frac{x+a+y+b-(x+y)}{a+b}=\frac{a+b}{a+b}=1$$
So did i do my math right. Also all my answers were 1, so what does that mean in relation to my posted question.
Thank you
You forgot that the denominator has $h$ in it - not $f(h)$ so for example $$\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}=\lim_{h\to 0}\frac{x+a+y+b-(x+y)}{a+bi}=\lim_{h\to 0}\frac{a+b}{a+bi}\ne1$$