# Always real valued complex function

Consider the function $f:\mathbb{C}\setminus\{a,b,c,d\}\to\mathbb{C}$ defined by $$f(z)=\dfrac{z}{z-a}- \left(\dfrac{z}{z-b}+\dfrac{z}{z-c}+\dfrac{z}{z-d}\right).$$ I,m trying to find the relationship between $a,b,c,d$ so that $f$ always assumes only real values.

How can I find such a condition?
Thank you.

• $a=b=c=d=0$ ?  – Amr Dec 18 '15 at 8:51
• Put $z = x+ iy$, evaluate $f(z)$. Now you can find the relation from the equation $\Im(f(z)) = 0$ – user297008 Dec 18 '15 at 8:51
• Do you mean real values for real arguments? Otherwise f can only be constant. – Martin R Dec 18 '15 at 8:54
• $z$ was arbitrary. – user297008 Dec 18 '15 at 8:54
• Your function is analytic on its domain. It’s well-known that an analytic function that takes only real values must be constant. Why? Analytic functions are open mappings, i.e. must map open subsets of the domain to open subsets of the target space. – Lubin Dec 19 '15 at 4:09

Your function is analytic on its domain. It’s well-known that an analytic function that takes only real values must be constant. Why? Nonconstant analytic functions are open mappings, i.e. must map open subsets of the domain to open subsets of the target space.

Hint:

Step-1 Simply the expression- $\frac{z}{z-a}=1+\frac{a}{z-a}$. Do it for all.

Step-2 Make the denominator real- $\frac{1}{z-a}=\frac{\bar{z}-\bar{a}}{\|z-a\|^2}$. Do it for all.

Step-3 Do $f(z)-\bar{f(z)}$ and find when it is zero.

• In Step-1, it should be ${z \over z-a}$ – user297008 Dec 18 '15 at 9:02
• Yes, that is true. – Rajat Dec 18 '15 at 11:03
• Resulting expression wont be independent from $z$. – Bumblebee Dec 19 '15 at 3:59
• But you will find it as the following form $\sum_{i=1}^{k}a_i(z-\bar{z})$ where $a_i$s are real numbers. – Rajat Dec 19 '15 at 11:20