Fiding imaginary part of a complex number

Question

What is the imaginary part of $i^i$ ?

I've tried multiple approaches, including using log. I can't seem to understand how to work with complex numbers as logarithmic functions. Also, it would help if someone could explain what the value of $\log{(-i)}$ is?

Thanks in advance :)

$i^i$ is multivalued, and always real. The imaginary part is zero. — Deepak, Apr 24 '16 at 13:57

score 1 · Accepted Answer · answered Apr 24 '16 at 14:08

1

You can write $i^i=e^{i\log i}$ for some suitable value of $\log i$. One fine choice is $\log i = \frac{\pi i}{2}$. In that case, we obtain:

$i^i = e^{i\cdot\frac{\pi i}{2}} = e^{-\pi/2}$,

which is a real number with imaginary part $0$. Indeed, any choice for $\log i$ is going to be a purely imaginary number, so when you multiply it by $i$, you'll get a purely real number.

Does that make sense?

answered Apr 24 '16 at 14:08

G Tony Jacobs

31,218

Thank you so much, it makes perfect sense – b_boundary Apr 24 '16 at 14:09

score 1 · Answer 2 · edited Apr 13 '17 at 12:21

1

$i^i$ is real and $=e^{-\pi/2} \approx0.2078795\dots$ thus its imaginary part is $0$.

I believe a similar Question about the value of $i^i$ was asked before, so you can see that for more information.

You can also see simple Mathematical Proof Using the Euler’s Formula.

edited Apr 13 '17 at 12:21

Community

1

answered Apr 24 '16 at 14:14

Vepir

12,516

Fiding imaginary part of a complex number

2 Answers2