Why is the derivative of i 0

Question

I would like to know why the derivative of $i$ is 0. Someone explained that $i$ is a constant, but I thought $i$ was a complex number.

Also, is the derivative of, for instance, $5i$ = 0? Also, what would be the derivative of $2+3i$?

Thank you!

Answer 1

Compare the derivatives of real and complex functions: $$f(x)=3 \Rightarrow f'(5)=\lim_{h\to 0} \frac{f(5+h)-f(5)}{h}=\lim_{h\to 0} \frac{3-3}{h}=0\\ f(x)=x \Rightarrow f'(5)=\lim_{h\to 0} \frac{f(5+h)-f(5)}{h}=\lim_{h\to 0} \frac{(5+h)-5}{h}=1\\ --------------------------------------\\ f(z)=i \Rightarrow f'(5)=\lim_{h\to 0} \frac{f(5+h)-f(5)}{h}=\lim_{h\to 0} \frac{i-i}{h}=0\\ f(z)=z+2 \Rightarrow f'(5)=\lim_{h\to 0} \frac{f(5+h)-f(5)}{h}=\lim_{h\to 0} \frac{(5+h+2)-(5+2)}{h}=1$$ Note:

1) $f(z)=i \Rightarrow f(2)=f(-3)=f(2+3i)=i$.

2) The derivative of a function shows its rate of change and a constant function does not change, hence its derivative is zero.

Answer 2

The derivative of any complex constant is $0$. This follows from the definition of (complex) derivatives: $$ f'(z_0)=\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}. $$

For instance, suppose $f(z)=3+5i$ for all $z$. Then the limit above would be $0$ for any $z_0$.

Answer 3

$i$ is a constant like any other number

Now you say that it is a complex number .Yeah it's true but it is a constant number. When i change the value of $x$ value of $i$ remains the same. That's why it is a constant . $5i$ is a constant because whatever value i put in x the value of 5i remains the same. Like any other number i is also a number. What you are saying is like "why is 2 a constant, it is an integer"

Derivative of 2+3i in respect to x would be 0 because whatever value I input in x value of 2+3i will never change