# How to Compute Complex Exponents (a+bi)^{c+di}

Any complex number *a + bi* can be written in the equivalent forms*re ^{iθ}* or

*r*[cos(

*θ*) +

*i*sin(

*θ*)],

where

*r*is the modulus (length) of

*a + bi*and

*θ*is the argument (angle) of

*a + bi*in the complex plane, measured in radians. You can find

*r*with the formula

*r*= sqrt(

*a*²+

*b*²)

and

*θ*with the formula

*θ*= arctan(

*b*/

*a*) +

*K*π,

where

*K*is a non-unique integer that depends on the signs of

*a*and

*b*.

To simplify an exponential expression of the form (

*a + bi*)

^{c + di}, first rewrite the expression as

*e*

^{(c + di)LN(a + bi)}. Then simplifiy it using properties of exponents and logarithms as follows:

*e*

^{(c + di)LN(a + bi)}

=

*e*

^{(c + di)LN(reiθ)}

=

*e*

^{(c + di)[LN(r) + iθ]}

=

*e*

^{[cLN(r) - dθ] + [dLN(r) + cθ]i}

This can be split into the product of two exponentials, one with a real number power and one with an imaginary number power:

(

*e*

^{cLN(r) - dθ})(

*e*

^{[dLN(r) + cθ]i})

The first factor is a real number equal to the modulus of the complex number and the second factor contains the new angle,

*d*LN(

*r*) +

*cθ*. The final answer can be written in standard form

*x + iy*:

*e*

^{cLN(r) - dθ}cos(

*d*LN(

*r*) +

*cθ*) +

*i*

*e*

^{cLN(r) - dθ}sin(

*d*LN(

*r*) +

*cθ*)

### Non-Uniqueness of Final Answers

Because any angle*θ*is equivalent to

*θ*+ 2

*n*π for any integer

*n*, there are infinitely many ways to write the angle of

*a + bi*and thus infinitely many complex numbers that equal (

*a + bi*)

^{c + di}. In most practical applications one takes the unique value of

*θ*that lies in the interval [0, 2π) or the interval (-π, π].

### Special Values

*i*=

^{i}*e*

^{iLN(eiπ/2)}=

*e*

^{-π/2}

(1 +

*i*)

^{1 + i}=

*e*

^{(1 + i)LN(√2eiπ/4)}= (

*e*

^{LN√2 - π/4})(

*e*

^{[LN√2 + π/4]i})

© *Had2Know 2010
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