# Thread: [SOLVED] Proof of the chain rule on holomorphic function

1. ## [SOLVED] Proof of the chain rule on holomorphic function

Okay. A) This just doesn't work. That's a pretty huge problem.

B (probably the answer to the problem))

We know any function on a complex variable can be decomposed:

$\displaystyle f(z) = Re(f(z)) + i[Im(f(z))] = u + iv$

Where u(x,y) and v(x,y) are real-valued functions.

Now, let $\displaystyle g(z) = s + it$ similarly.

Thus, $\displaystyle g(f(z)) = s(u(x,y) + i[v(x,y)]) + i[t(u(x,y) + i[v(x,y)])]$.

so $\displaystyle (g(f(z))' = (s(u(x,y) + i[v(x,y)]))' + (i[t(u(x,y) + i[v(x,y)])])'$

$\displaystyle = s'(u(x,y))u'(x,y) + s'(i[v(x,y)])iv'(x,y) + i[t'(u(x,y))u'(x,y) + t'(i[v(x,y)])iv'(x,y)$

Right? But then what happens to the i's inside the real-valued functions?

2. Not right! Clearly there should be no imaginary numbers inside your real-valued functions. If $\displaystyle f(z) = f(x,y)=u(x,y)+iv(x,y),\, g(z)= g(x,y)=s(x,y)+it(x,y)$, then $\displaystyle f(g(z)) = f(s,t)=u(s,t)+iv(s,t)=u(s(x,y),t(x,y))+iv(s(x,y),t (x,y))$. See if that helps!

3. This is what I was thinking, but I wasn't sure if I could do that. Thank you!

4. Originally Posted by davismj
This is what I was thinking, but I wasn't sure if I could do that. Thank you!
You're welcome! But make sure you understand why what I wrote makes sense. It's just basic function composition! Think about it and report back if you're still not sure.