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### Complex Exponentiation

```Date: 02/20/2015 at 19:27:47
From: Jacob
Subject: Raising complex numbers to/and complex powers

Okay, I have a "biggie" for you.

I am trying (and failing) to raise a complex number to a complex power.

Actually, the different cases involving complex powers with and without
base-e have been scattered across your archives, so I was thinking I'd
make it easier for the next person to find an answer all in one place.
(Also, I'm not really sure I'm doing half of them right.) Would you mind
explaining all four cases?

I know Euler's Identity, (e^(xi) = cos(x) + isin(x)), and the natural
logarithm identity a = e^(ln(a)). But I have been looking for MANY hours
on how to do this (mainly the last one), and I cannot find a decent
explanation of how to do it by hand (well, not COMPLETELY by hand...).

But mostly, I'm having trouble solving x^y when BOTH x and y are complex
numbers.

Okay, here are my four example cases.

I DO think that I got this one right:

e^(3i)
= cos(3) + isin(3)
= .989992 + .141120i

I think that I got this one right, too:

e^(2 + 3i)
= e^2*e^(3i)
= e^2*cos(3) + isin(3)
= -7.315110 + .141120i

I'm not sure if I got this one right:

6^(2 + 3i)
= e^(ln(6^(2 + 3i)))
= e^((2 + 3i)ln(6))
= e^(2ln(6))*e^(3i)
= 36*cos(3) + isin(3)
= -35.639729 + .141120i

Here's the one that I really can't seem to figure out:

(2 + 3i)^(3 + 2i)
= e^(ln((2 + 3i)^(3 + 2i)))
= e^((3 + 2i)ln(2 + 3i))
= e^((3 + 2i)^(ln(2 + 3i))) ...

And that's where I can't continue.

I know you can't split it into its respective parts, because you're
raising the WHOLE base (in this case, 3 + 2i) to the power (in this one,
ln(2 + 3i)).

The most difficult part about these, I think, is the fact that you're
breaking apart everything you've learned. I guess we have Euler to thank
for making it even possible; but multiplying a number by itself a non-real
number of times already strains the intuition.

Can you help me out here? For the life of me, I cannot understand this
one. I'd appreciate it forever.

I'm one of those people who can't sleep until they figure out the problem
that they've started.

Any help at all is appreciated! Thank you!!

```

```
Date: 02/20/2015 at 22:14:48
From: Doctor Ali
Subject: Re: Raising complex numbers to/and complex powers

Hi Jacob!

Thanks for writing to Dr. Math.

I think the following answer from the Dr. Math archives will tell you what
you want to know:

http://mathforum.org/library/drmath/view/52251.html

I found it by searching on these keywords:

complex power

You can find other similar answers by doing the same search at

http://mathforum.org/library/drmath/mathgrepform.html

or if you have any other questions.

Please write back if you still have any difficulties.

- Doctor Ali, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 02/22/2015 at 16:38:20
From: Jacob
Subject: Raising complex numbers to/and complex powers

Dear Dr. Ali,

Thank you so much for your help!

Unfortunately, I still don't completely understand a complex power raised
to a complex power.... Would it be possible to explain it a little more in
depth for me? Thank you!!

Jacob

```

```
Date: 02/23/2015 at 23:30:34
From: Doctor Ali
Subject: Re: Raising complex numbers to/and complex powers

Hi Jacob!

Thanks for writing to Dr. Math.

Let's say that we want to find z1^z2, where

z1 = a + b i
z2 = c + d i

This is the most general case in raising complex numbers to complex
powers. We can write it as

(a + bi)^(c + di) = (a + bi)^c x (a + bi)^(di)

The first part, (a + bi)^c, is pretty easy. If c is an integer, you'll
just need to use the binomial expansion and expand (a + bi)^c and do the
calculation, bearing in mind that i^2 = -1. If it's not an integer, you'll
need to write (a + bi) in polar form and use De Moivre's Formula to find
the answer in polar form. That is,

a + bi = r exp(i t)

So,

(a + bi)^c = (r exp(i t))^c = r^c exp(i t c)

It seems you don't have a problem with the first part, but in case you

http://mathforum.org/library/drmath/sets/select/dm_demoivre.html
http://en.wikipedia.org/wiki/De_Moivre's_formula

Now for the second part, we want to exaluate (a + bi)^(di). We can write
it as

(a + bi)^(di) = ((a + bi)^d)^i

The term (a + bi)^d is just like above. Now, you have a complex number
that you want to raise to i-th power. You can easily do this using the
same method discussed in the below page from our archive:

http://mathforum.org/library/drmath/view/53860.html

I will let you do this part yourself as an exercise. What you are supposed
to do is to find M and N in the expression below:

(x + yi)^i = M + N i

Once you master this technique, you'll see that you can directly separate
the real part and imaginary part of the power in the original question and
use De Moivre's Formula to find the result of each.

Please write back if you still have any difficulties.

- Doctor Ali, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 02/26/2015 at 10:25:09
From: Jacob
Subject: Raising complex numbers to/and complex powers

Hi Dr. Ali,

I apologize for the late reply. I've had a lot going on.

I did have a chance to check out the links, and they explain a lot to me.
However, I still am not sure how to get past a certain part in solving

(2 + 3i)^(3 + 2i)

I can get this far:

-46 + 9i(-5 + 12i)^i

But from there, I have no idea where to go. I didn't see that they did it
in those other answered questions, either. I know you can't "distribute"
the exponent (in this case, i), and I don't think you can make a logarithm
base -5 + 12i of (x/(-46 - 9i)).

And I obviously can't write out and multiply -5 + 12i an i number of times....

This is driving me crazy!

Thanks for all of your help!

Jacob

PS. I saw in another article that you can take the i index of a number
(basically, raise it to a (1/i) power). I don't really understand how they
got the answers that they did. Can you explain that to me (or send me a
link with the work written out)?

I'm so sorry, I've created more questions for you.

Thank you so much!!

```

```
Date: 02/26/2015 at 23:05:32
From: Doctor Peterson
Subject: Re: Raising complex numbers to/and complex powers

Hi, Jacob.

Here is what I would do to find (2 + 3i)^(3 + 2i).

We first have to express 2 + 3i in polar form. We find that the angle
(argument) is arctan(3/2), which I will just call A for now to simplify my
work, and the length (modulus) is sqrt(2^2 + 3^2) = sqrt(13). Thus,

2 + 3i = sqrt(13))*e^(Ai)
= e^(ln(sqrt(13)) + Ai)

Let's use R to represent ln(sqrt(13)), so

2 + 3i = e^(R + Ai)

Now,

(2 + 3i)^(3 + 2i) = (e^(R + Ai))^(3 + 2i)
= e^((R + Ai)(3 + 2i))
= e^((3R-2A) + (2R + 3A)i)
= e^(3R-2A)*e^(2R + 3A)i)
= e^(3R-2A)[cos(2R + 3A) + sin(2R + 3A)i]
= e^(3R-2A)cos(2R + 3A) + e^(3R-2A)sin(2R + 3A)i

Numerically,

A = arctan(3/2)  = 0.98279
R = ln(sqrt(13)) = 1.28247

3R - 2A = 1.88183       e^(3R - 2A) = 6.5655
2R + 3A = 5.51333      cos(2R + 3A) = 0.718012
sin(2R + 3A) = -0.6960310

So our result is

6.5655*0.718012 + 6.5655*-0.6960310i = 4.714 - 4.569i

Does that give you a better sense of where an answer like this comes from?

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 03/08/2015 at 10:05:43
From: Jacob
Subject: Thank you (Raising complex numbers to/and complex powers)

Dr. Ali,

Again, sorry for the late response.

I get it now. Thank you so much! This is awesome!!

I really appreciate all that you have done for me. It means so much.

I wish you the best!

Happy Math!

Jacob
```
Associated Topics:
High School Exponents
High School Imaginary/Complex Numbers
High School Logs

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