# complex inputs function

#### ltn

##### New member
Hello!
I've encountered problem:

and best I came up with is:
t = 3 + ih
-ih = -t + 3
ih = t - 3
2 + 4(t-3)
4t - 10
f'(3) = 4???

I suppose, this shouldn't be solved that way, so my question is: how can this type of problem be solved?

#### Jomo

##### Elite Member
Let's see: 3 = 3+ih if h=0.
So f( 3) = f( 3+i*0) =....

#### LCKurtz

##### Full Member
Is [imath]h[/imath] real? If [imath]f[/imath] is a real valued function, how can $f(3+2ih) = 2+ 4ih$?

pka

#### Dr.Peterson

##### Elite Member
I searched for the problem to see if it has any context that would explain the idea of a real-valued function (for real inputs?) that can also have complex inputs and outputs; the problem is found here, which appears to be rather specialized (though I couldn't find the explanation I hoped to find):

My best guess is that you are expected to use the limit definition of the derivative, and hope it will be meaningful.

#### Jomo

##### Elite Member
Is [imath]h[/imath] real? If [imath]f[/imath] is a real valued function, how can $f(3+2ih) = 2+ 4ih$?
Good point. I missed that one.

#### ltn

##### New member
Hello!
Turns out this problem could be solved using this equation (link for reference)

So, eventually:

[imath]f'(x)=\frac{lm(2+4ih)}{h}= 4h/h = 4[/imath]

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