# To the power of an imaginary

1. Mar 17, 2007

### Jeff Cook

All,

I understand by rules of complex math, that raising a real number to the power of a complex number, you simply drop the imaginary part; it is not affected at all. But what happens when you raise a real number to the power of an imaginary. For instance...

x = 2^3i - 2

Does x end up equal to zero or negative 2 in this example? Or something else altogether?

Any help with this would be greatly appreciated.

Thanks,

Jeff

2. Mar 17, 2007

### Jeff Cook

Got it from John Derbyshire in his book, "Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics." On page 333...

"If you raise a real number x to a complex power a + bi, the rules of complex arithmatic dictate the following. The modulus of the answer-how far it is from zero, as the crow flies--is x^a. It is not affected by b at all...the amplitude, however, might be anything at all."

Can you better explain?

Jeff

So the value of the

3. Mar 17, 2007

### matt grime

Complex numbers are best written in exponential form

$$x+iy = re^{i\theta}$$

then it is straight forward to see what's going on.

4. Mar 17, 2007

### D H

Staff Emeritus
Complex exponentials, like real exponentials, obey the relationship
$$\exp(a+b) = \exp(a)\cdot\exp(b)$$
Thus
$$\exp(a+bi) = \exp(a)\cdot\exp(bi)$$
The former term is a real number. Using Euler's formula on the latter term,
$$\exp(bi) = \cos b+i\sin b$$
which is a point on the unit circle (i.e., its magnitude is one). Thus the magnitude of the complex exponential $\exp(a+bi)$ is simply $\exp a$. However, the amplitude cannot be anything at all. The answer is very well defined.

5. Mar 17, 2007

### Jeff Cook

Okay, that makes things very clear...So, in my example...

x = 2^3i - 2

Where we have two real numbers, but one raised to an imaginary. x = -1? Is this correct?

Jeff

6. Mar 17, 2007

### D H

Staff Emeritus

Hint: With real numbers, one can use
$$a^b = \exp(b\log a)$$
This works with complex exponentiation as well.

7. Mar 17, 2007

### Jeff Cook

Well, the answer negative one was a misunderstanding on my part. I have a new answer and if it's wrong, I hope you will be able to walk me through to the correct equation.

2^3i = exp(3i log(2))
= (1/8)i

X = (1/8)i - 2
= -2 + (1/8)i
= -2.125

Is that correct?

Jeff

8. Mar 17, 2007

### Jeff Cook

Oops, I forgot something important. Hold on one minute.

Jeff

9. Mar 17, 2007

### Jeff Cook

2^3i = exp(3i log(2))
= exp(2.079i)
= cos (2.079) + i sin (2.079)
= -.487 + .873i
= .386...

Is that correct?

Jeff

10. Mar 17, 2007

### Jeff Cook

And...

.386 - 2 = 1.614 = x

11. Mar 18, 2007

### Mute

Not quite correct. You are correct in stating

2^(3i) = -0.487 + 0.873i, but the following line,
= 0.386...

is not correct. 2^(3i) = -0.487 + 0.873i (to 3 decimals of precision) is the answer, and this answer is a complex number. A complex number, z, always consists of a "real part" and an "imaginary part" (note that this is just terminology): z = x + iy; x is the real part of z and y is the imaginary part. i is the imaginary unit such that i^2 = -1, of course. You can no more add the real and imaginary parts of a complex number than you can add the x and y component of a vector.

x = 2^(3i) - 2 evaluates as
x = -0.487 + 0.873i - 2 = -2.487 + 0.873i

(Also, there is a slight subtlety in all of this, in that in stating some complex number z = exp(log(z)), you need to specify the "branch" of log(z) to make your result single valued. If we write $z = r \exp(i \theta)$, then $\log(z) = \ln r + i \arg(z)$, where $\arg(z) = \theta + 2 \pi k$, for any integer k. You need to specify that integer k, otherwise you will get infinitely many results for your answer. The answer above used k = 0.)

Last edited: Mar 18, 2007
12. Mar 19, 2007

### Jeff Cook

Mute,

Thanks. I understand that part of it now. But I still have then a couple of other questions that come up. I will add a different example to better clarify things for me.

In the equation:

ln((-1-2za)+ -i) = (.5+0i) ln((-1 + Sqrt{8}) + 2i)

How would one go about solving for z, a real part of a complex number in an equation consisting entirely of complex numbers. I added the 0i in order to make the real numbers fully understandable as complex numbers.

Can you, or someone, help me now with this example. It is the source of my problem in getting this straight.

Thanks greatly!

Jeff

13. Mar 19, 2007

### HallsofIvy

Staff Emeritus
You'd start in exactly the same way you would with real numbers- get rid of the logarithms:
$$ln(-1-2za- i)= ln(\sqrt{-1+ \sqrt{8}+ 2i})$$
so
$$-1- 2za- i= \sqrt{-1+ \sqrt{8}+ 2i}[/itex] Squaring both sides, [tex]4za+ 4z^2a^2+ 4zai= -1+ \sqrt{8}+ 2i$$
$$4a^2 z^2+ (4a+ 4ai)z+ (1- \sqrt{8}-2i)= 0$$
Now solve that with the quadratic formula.

14. Mar 19, 2007

### Gib Z

Perhaps one should be careful of an extra solution?

15. Mar 19, 2007

### Jeff Cook

I'm not sure if using a quadratic equation is the best idea for this particular equation, as a = (1 / Sqrt {2})-1. This leads simply to one quadratic equation after another, on and on. I'll show.

log ((-1-2za) + -i) = (.5 + 0i) log ((- 2 + Sqrt {8}) + 2i)

0 = z^2 (4a^2) + z (4a + 4ai) + (1 - Sqrt {8} - 2i)

t = 4a^2

u = 4a + 4ai

v = 1 - Sqrt {8} - 2i

z = 1.7071068 - .8040405i +/- Sqrt {-3.5492065 + 5.4903320i)} / .6862915

I'll just show you one of the paths from here…

(.6862915 ((z +0i) - (1.7071068 - .8040405i)))^2 = -3.5492065 + 5.4903320i

And so I don't need to write out all the numbers…

f = .6862915
g = 1.7071068
h = -.8040405i

(f (2z-2g) -2h)^2 = -3.5492065 + 5.4903320i

((1.3725830z - 2.3431458) + 1.1036123i)^2 = -3.5492065 + 5.4903320i

Again the numbers are too long for me to write out, so…

j = 1.3725830
k = 2.3431458
l = 1.1036123

((jz - k) + li) (jz - k) + li) = -3.5492065 + 5.4903320i

(jz - k)^2 + 2li (jz - k) -1.2179602 = -3.5492065 + 5.4903320i

((jz - k)^2 - 1.2179602) + 2li (jz - k) = -3.5492065 + 5.4903320i

Which I believe by looking at it leads to another quadratic equation. Correct me if I am wrong. If you can solve z as a complex number going in this direction, please show me.

However, I took a different approach and went in the opposite direction. Starting again with the first equation…

log ((-1-2za) + -i) = (.5 + 0i) log ((- 2 + Sqrt {8}) + 2i)

Could you please check my math in what follows?

log (-1+ -i - 2za + 0i) = (.5 + 0i) log ((- 2 + Sqrt {8}) + 2i)

I assume for a moment that this came from a quadratic equation, as taking it in the opposite direction leads to many other quadratic equations. So…

-1+ -i + 2za + 0i = (- 2 + Sqrt {8} + 2i)^.5

or…

-1+ -i - 2za + 0i = (- 2 + Sqrt {8} + 2i)^.5

So,

2za + 0i = (-1+ -i) +/- Sqrt {- 2 + Sqrt {8} + 2i}

z + 0i * 2a + 0i = (-1+ -i) +/- Sqrt {- 2 + Sqrt {8} + 2i}

z = ((-1 + i) +/- Sqrt {-2 + Sqrt {8} + 2i}) / (2a + 0i)

z = ((-1 -i) +/- Sqrt {2i - 2 + Sqrt {8}}) / 2a

Now, because of the following…

(1+i)(1+i) = 1 + i + i -1 = 1 + 2i - 1 = 2i

z = (-b +/- Sqrt{b^2 - 2 + Sqrt {8}}) / 2a

Where…

a = (1 / Sqrt {2}) - 1

b = 1 + i

c = 1 / Sqrt {2}

Thus,

0 = z^2 ((1 / Sqrt {2}) + i^2) + -z (-1 - i) + 1 / Sqrt {2}

Rearrange…

z (-1 - i) - 1 / Sqrt {2} = z^2 (1 / Sqrt {2} + i^2)

-z - zi = 1 / Sqrt {2} + z^2 / Sqrt {2} + z^2i^2

-1 - i = 1 / zi Sqrt {2} + z / Sqrt {2} + zi^2

Divide both sides by i^2.

1 + i = zi / Sqrt {2} + zi^2 / Sqrt {2} + z

Divide both sides by i.

1 / -i + 1 = z / Sqrt {2} + zi / Sqrt {2} - zi

1 / -zi + 1 / z + i = 1 / Sqrt {2} + i / Sqrt {2}

1 / -zi + 1 / z + i = Sqrt {i)

i / z + 1 / z + i = Sqrt {i}

i / z + 1 / z = Sqrt {i} - i

i / z = -1 - Sqrt {i} + 1 / Sqrt {2} - I

Lastly, I get what appears as an identity of z, where z could be any number of possibilities if taken the other direction. However, in this direction, I get…

z = i (-1 - i / Sqrt {2} - i)^-1

But what does that equal? How to reduce this further? I am stuck here. The best I get is z = 0. But I have already added enough of my work and do not know if I have made it correctly even this far. Could someone help me out with this?

Thanks greatly!

Jeff