## Basic power rule ((a^b)^c = a^(b*c)) does not work

### Andreas Dorner (view profile)

on 1 Feb 2019
Latest activity Commented on by Luna

on 4 Feb 2019

### Luna (view profile)

Just cracked down a large problem in my code to a strange phenomenon. A very simple power rule doesn't seem to work here.
b = 5i; %complex, this should have something to do with it
c = 0.1; %for intergers it seems to work
x1 = exp(b)^c;
x2 = exp(b*c);
x1 =
0.9918 - 0.1280i
x2 =
0.8776 + 0.4794i
Shouldn't they be equal?
I just do not unterstand..

Greg Dionne

### Greg Dionne (view profile)

on 1 Feb 2019
I can't tell if this is homework or not, so here's a thought experiment:
At what value of x does the following print different values for your two expressions?
for x=0.01:.01:5;
y = 0.5/x; % make sure x*y = 0.5
expression1 = exp(1i*x)^y;
expression2 = exp(1i*x*y);
fprintf('%f: %f + %fi ?= %f + %fi\n',x, ...
real(expression1), imag(expression1), ...
real(expression2), imag(expression2));
end
Andreas Dorner

### Andreas Dorner (view profile)

on 1 Feb 2019
No, it's not homework, this is was a problem i encountered in academic research. I wanted to get rid of one power to speed up my code running on gpu.
I see where you're going with this, thank you! exp(1i*2pi) is starting over again at the unit circle. I didn't see that coming, i feel kinda stupid now. But I wonder if a can use that knowledge somehow.
Thank you very much for your insight. Have a nice day!

R2018b

on 1 Feb 2019
Edited by Luna

### Luna (view profile)

on 1 Feb 2019

By the definition Euler's equation: z = x + iy means
So the two things might be different:
and
x2 = exp(b*c) is equal to exp(c)^b actually.
equals
equals also equals
But those two above are not equal.

Walter Roberson

### Walter Roberson (view profile)

on 1 Feb 2019
(I notice the latex is cut off at the top of each section. I have reported this to Mathworks a second ago.)
Luna

### Luna (view profile)

on 4 Feb 2019
Yes, it seems an issue :) Thanks!

### James Tursa (view profile)

on 1 Feb 2019
Edited by James Tursa

### James Tursa (view profile)

on 1 Feb 2019

This has been discussed in this forum before. Raising complex numbers to a power is a multi-valued operation. MATLAB picks one of those results, which may or may not agree with the result you get by rearranging the operations. E.g.,
>> b = 5i;
>> c = 0.1;
>> exp(b)
ans =
0.283662185463226 - 0.958924274663138i
>> exp(b+2*pi*i) % You can add 2*pi to get same result
ans =
0.283662185463226 - 0.958924274663139i
>> exp(b+4*pi*i) % You can add 4*pi to get same result
ans =
0.283662185463226 - 0.958924274663139i
>> exp(b+18*pi*i) % In general, you can add k*2*pi
ans =
0.283662185463224 - 0.958924274663139i
>>
>> exp(b)^c
ans =
0.991778467700342 - 0.127966679280045i % The MATLAB answer for this expression
>> exp(b*c)
ans =
0.877582561890373 + 0.479425538604203i % This particular rearrangement doesn't match
>> exp((b+18*pi*i)*c)
ans =
0.991778467700342 - 0.127966679280045i % But this one does
You can probably look at the doc to see the rules for how MATLAB picks the result for these mutli-valued issues. But the bottom line is if you have code that depends on multi-valued calculations like this, you need to account for it in your code logic. MATLAB isn't going to know which one you want picked, and you shouldn't expect it to.

Andreas Dorner

### Andreas Dorner (view profile)

on 2 Feb 2019
Thank you first of all. I see the problem.
But why does exp((b+18*pi*i)*c) work? 18pi seems a bit arbitrary
Stephen Cobeldick

### Stephen Cobeldick (view profile)

on 2 Feb 2019
"18pi seems a bit arbitrary"
It is entirely arbitrary. That is exactly the point.