# How do I find $\frac{d}{dz}\left(\frac{2z-i}{z+2i}\right)\text{?}$

How do I find:

$$\frac{d}{dz}\left(\frac{2z-i}{z+2i}\right)\text{?}\quad\quad z\in\Bbb C$$

Do I turn it into an $x+iy$ form and use the Cauchy-Riemann equations?

I couldn't get it into such a form, so I tried the Quotitent rule, treating $z$ as if it were real, so pretty much just calculus - I got the answer slightly wrong(probably a little error somewhere), but it took so many steps that I doubt this is the correct method.

• The quotient rule works perfectly and is very very simple. Actually, all the usual derivation rules hold ($+,-,\times,\div,f^{-1}(z),g(f(z)),z^n,e^z\cdots$). Just take care in case of non holomorphic terms, like $|z|$.
– user65203
Commented Apr 20, 2015 at 9:37

For the numerator, we have $$2z-i=2x+2yi-i=2x+(2y-1)i$$ So now $$\frac{\partial}{\partial x}[2x] =\frac{\partial}{\partial y}[2y-1]= 2$$ And $$\frac{\partial}{\partial x}[2y-1] =-\frac{\partial}{\partial y}[2x]= 0$$ Therefore $$2z-i\Rightarrow \mbox{is analytic}$$ For the denominator, we have $$z+2i=x+yi +2i=x+(y+2)i$$ So now $$\frac{\partial}{\partial x}[x] =\frac{\partial}{\partial y}[y+2]= 1$$ And $$\frac{\partial}{\partial x}[y+2] =-\frac{\partial}{\partial y}[x]= 0$$ Therefore $$z+2i\Rightarrow \mbox{is analytic}$$ Since $2z-i$ and $z+2i$ are both analytic, then $$\frac{2z-i}{z+2i}\Rightarrow \mbox{is analytic}$$ Hence $$\frac{d}{dz}\left[\frac{2z-i}{z+2i}\right]$$ $$=\frac{(z+2i)\frac{d}{dz}[2z-i]-(2z-i)\frac{d}{dz}[z+2i]}{(z+2i)^2}$$ $$=\frac{2z+4i-2z+i}{(z+2i)^2} =\frac{5i}{(z+2i)^2}$$

• Omg I am so sorry I mistyped the question, I don't know what to do now. It was meant to be $\frac{d}{dz}$ Commented Apr 20, 2015 at 7:55
• @SkiesBurn, you can simply just edit your question.
– k170
Commented Apr 20, 2015 at 7:56
• Just quotient ruled it? Would it be expected to use Cauchy-Riemann first to ensure it is analytic? Commented Apr 20, 2015 at 8:07
• @SkiesBurn, Please see my edit.
– k170
Commented Apr 20, 2015 at 9:26
• None of this is required to solve the question as it is now.
– Did
Commented Apr 26, 2015 at 12:50