# Principal value

1. May 31, 2014

### MissP.25_5

Hello.
Can someone please check if I got the answer right? I don't have the right answers for my exercise so I need someone to check it for me, please.

Please find the principle value of (1+i)^-i

My attempt:
√2^(- i) = (e^ln√2)^(- i) = e^-i•ln√2
= cos(- ln√2) + i•sin(- ln√2)
= cos(ln√2) - i•sin(ln√2)

(1 + i) = √2[(1/√2) + i(1/√2)]
= √2[cos(π/4) + isin(π/4)]
= √2•e^i(π/4)

(1 + i)^(- i) = [√2•e^i(π/4)]^(- i)
=[e^(π/4)] [√2^(- i)]
=[e^(π/4)][cos(ln√2) - i•sin(ln√2)]

Therefore (1 + i)^(- i) = [e^(π/4)][cos(ln√2) - i•sin(ln√2)]

2. Jun 1, 2014

### SteamKing

Staff Emeritus
It's not clear what these calculations represent:

You seem to have trouble expressing (1+i) in polar form and in applying Euler's formula correctly.

Remember:

z = (x+iy) = r(cos θ + i sin θ) = r e$^{i θ}$

where

r = $\sqrt{x^{2}+y^{2}}$

and

θ = arctan (y/x)

3. Jun 1, 2014

### MissP.25_5

But what's wrong? r=√2 and θ=∏/4.