What is the principal value of (1+i)^i ?

You should note that raising to an imaginary power is defined by

.. iΘ

e.....= cosΘ + i sin Θ

and

..a + iΘ.........a.....iΘ.........a

e.............= e....e.......= e (cosΘ + i sin Θ)

The famous special case is

..iπ

e.....= -1

so

......... i

(1 + i)

is analyzed by noting

1 + i

is a complex number with

....................._.....___

magnitude √2 =√1 +1

and direction 45° which is π/4

so it equals

√2 [ cos π/4 + i sin π/4 ]

Now we wish to raise this number to the i

power. i = i times 1

Let us convert EVERYTHING to powers of e

............(ln2)/2

√2 = e

.........................................I(π/4)

cos (π/4 )+i sin (π/4) = e

Now

.........i

(1 + i)

equals

.........................................i

[√2 [ cos π/4+ i sin π/4 ] ]

equals

......I..............................i

(√2) (cosπ/4+ i sinπ/4 ]

equals

..........i(ln2)/2......i²(π/4)

= e...................e

.....-π/4......iln2/2

= e...........e

Approximate value for

e^( -π/4) is 0.455938128

....i(½ln2)

e = cos (ln2/2) + I sin(ln2/2)

arguments of cos and sin are in radians

These calculate to

approximately

0.940542105 + 0.339677125 i

Multiply the above expression by

..-π/4

e which is approximately 0.455938128

and we obtain

= 0.428829006 + 0.154871752 i

matching the calculations of Wolfram Alpha and other utilities.

Additional Note:

This value is the principal value as it is the ONLY {or unique} value for

.......i

(1+i)

There s a shorter way to get the principal value here. z^w for complex numbers z,w is actually defined as exp[w log(z)]. The complex exp function is single-valued:

exp(u + iv) = (e^u)(cos v + i sin v)

....where e^u, cos v and sin v are the usual functions of real numbers.

The principal value for z^w is defined as that expression using the principal value for log(z). Namely:

log(z) = ln(|z|) + i Arg(z)

... where ln(|z|) is the natural logarithm of the real number |z|, and Arg(z) is the "argument" angle of z in polar form, in the interval (-π, π].

Knowing that, the answer can be found with:

Arg(1+i) = π/4

|1 + i| = √2

log(1+i) = ln(√2) + iπ/4 = (ln 2)/2 + iπ/4

(1 + i)^i = exp(i log(1 + i)) = exp( i*[(ln 2)/2 + iπ/4]

The rest is simplification.

= exp(-π/4 + i[ln 2]/2)

= e^(-π/4)*{cos [ln(2) / 2] + i sin [ln(2) / 2]}

On a calculator, you ll get the approximations given, but this is exact. Another form might be:

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

...to save keystrokes on a calculator with a square root button.

0.428829006294368+0.154871752464247i

..although I don't know how it was calculated

see http://www.mathsisfun.com/numbers/complex-number-c...