Can complex number cancel out, as in $\frac{(3+2i)\cdot(1+i)}{(1+i)} = 3+2i$? If so, can I use that idea in complex differentiability?

Question

Can complex number cancel out? For instance, is the following statement true? $$\frac{(3+2i)\cdot(1+i)}{(1+i)} = 3+2i$$

If so, given a complex function $f:C\to C$, if I were to determine if it is complex differentiable (if $$\lim_{h \to 0} \frac{f(z_0+h)-f(z_0)}{h}= a+bi$$ for some $a,b\in R$, where $h$ is a complex number, then it is), could I cancel out the $h$ appearing in the numerator and denominator for any complex polynomial in the same method as performing first principle on real numbers?

Edit to clarify: For instance, for $f(z) = z^2$, I have $$ \lim_{h \to 0} \frac{(z+h)^2-z^2}{h}= \lim_{h \to 0} \frac{z^2+2z\cdot h-h^2-z^2}{h}= \lim_{h \to 0} \frac{2z\cdot h-h^2}{h}= 2z+\lim_{h \to 0} h=2z $$

Answer 1

$\frac {f(z+h) - f(z)}{h}$

$h$ has a real part and a complex part. i.e. $h = x+iy$

Similarly $f$ has a real and complex part.

$f(z) = u(z)+ iv(z)$

Making these substitutions our difference quotient becomes:

$\frac {u(z+x+iy) + iv(z + x+iy) - u(z)- i v(z)}{x+iy}$

We could rationalize the denominator by multiplying by the conjugate of $h,$ but it makes things a little messier than they need to be.

$\frac {xu(z+x+iy) - xu(z) + yv(z+x+iy) -yv(z) - ixu(z+x+iy) + iu(z) + iyv(z+x+iy) - iyv(z)}{|h|^2}$

Instead, suppose we consider small movements in the real direction separately from movements in the y direction. i.e. let $x$ be small while setting $y=0$

$\lim_\limits {x\to 0} \frac {u(z+x) - u(z)}{x} + i\frac {v(z+x) - v(z)}{x} = \frac {\partial u}{\partial x} + i\frac {\partial v}{\partial x}$

And then consider small movements in the imaginary direction (setting $x$ equal to $0$, and $y$ small but non-zero).

$\lim_\limits {y\to 0}\frac {u(z+iy) -u(z)}{iy} + \frac {iv(z+iy) -iv(z)}{iy} = \frac {1}{i} \frac {\partial u}{\partial y} + \frac {\partial v}{\partial y} = \frac {\partial v}{\partial y} - i\frac {\partial u}{\partial y}$

And a general movement of $h$ is just a combination of movements in the real part and the imaginary part.

Here is the real trick, though.

If the function is complex differentiable, then the derivative is the same for all small movements in $h$!

$\frac {\partial u}{\partial x} + i\frac {\partial v}{\partial x} = \frac {\partial v}{\partial y} - i\frac {\partial u}{\partial y}$

IF they are equal the real parts are equal and the imaginary parts are equal

$\frac {\partial u}{\partial x} = \frac {\partial v}{\partial y}\\ i\frac {\partial v}{\partial x} = - i\frac {\partial u}{\partial y}$

These are called the "Cauchy-Riemann equations." And a function is complex differentiable iff it satisfies these equations.

Example:

$f(z) = z^2 = (x+iy)^2 = x^2 - y^2 + i(2xy)\\ u(z) = x^2-y^2\\ v(z) = 2xy\\ \frac {\partial u}{\partial x} = 2x = \frac {\partial v}{\partial y}\\ \frac {\partial v}{\partial x} = 2y = -\frac {\partial u}{\partial y}$

All polynomials are complex differentiable, as are functions that can be represented as a polynomial. i.e. they have a Taylor Series, and you are evaluating within the radius of convergence of the Taylor series.

Answer 2

If $h$ is a complex number $a+bi$, $h$ can be cancelled out in the computation of the derivative so long as $a \neq 0$ and $b \neq 0$, i.e. $0+0i.$

Let's demonstrate what would happen if we didn't cancel the $1+i$; i.e we expand the denominator, and then multiply the numerator and denominator by the conjugate...

$$(3+2i)(1+i) = 3 + 5i - 2 = 1 + 5i$$

$$\dfrac {1+5i}{1+i} = \dfrac {(1+5i)(1-i)}{(1+i)(1-i)} = \dfrac {6+4i}{2} = 3+2i$$

Same number, just a little more work.

Answer 3

Since literally

$$ \frac{2z\cdot h-h^2}{h} = 2z - h $$

for any complex $h\neq 0,$ yes, you can "cancel" the $h$ as follows:

$$ \lim_{h \to 0} \frac{2z\cdot h-h^2}{h} = \lim_{h \to 0} (2z - h), $$

and then since each of the limits $\lim_{h \to 0} 2z$ and $\lim_{h \to 0} h$ is finite, you can proceed as you did in the question.