What are the rules for multiplication and exponentiation of complex numbers?

The other answers are absolutely right about how to do complex multiplication - just FOIL it, "pretending" that i is a variable. Once you're done FOILing, replace i^2 with -1.

Your question about "complex exponentiation" is...well...complex. If you're not in trig/pre-calc, but more like in Algebra I or II, I wouldn't worry too much about anything beyond (4 + i) ^3. To calculate that, you just do (4+i)(4+i)(4+i), and FOIL it.

In trig/pre-calc, or perhaps the tail end of a super-duper rigorous Algebra II course, you'll earn the "cis" form of complex numbers and DeMoivre's theorem, as well as the "complex roots of unity". If none of these sound familar, you can ignore the next paragraph:

To calculate the nth power of the number r * (cos theta + i * sin theta), you simply raise the distance to the nth power, and multiply the angle by n. Thus, you'd get:

r^n * (cos n*theta + i * sin n*theta)

There's a shorthand for the stuff in parentheses: cos theta + i *sin theta is often abbreviated as cis theta. (cis = cosine + i*sine, get it?) So you could write it as:

(r * cis theta)^n = (r^n) * cis (n*theta)

So if you had the number 2 * (cos 30 + i sin 30) and you wanted to raise it to the 6th power, you'd get:

2^6 * (cos 180 + i * sin 180) = 2^6 * (-1 + i * 0) = -64

As for complex numbers as EXPONENTS, like 5^i, it CAN be done, but is probably only covered in a complex analysis class that sophomores or junior math majors in college would take. Don't sweat it.

for (a + bi)(c + di) "foiling" it will give the correct answer:

(a + bi)(c + di) =

ac + adi + bci + bd(i^2) =

ac + (ad + bc)i - bd =

(ac - bd) + (ad + bc)i

or you can just memorize the formula:

(a + bi)(c + di) = (ac - bd) + (ad + bc)i

exponentiation is a little trickier. here's why:

it's not the "upper number" that gives so much trouble, it turns out that:

e^(c + di) = e^c(e^(di)) = e^c(cos(arctan(d/c)) + i sin(arctan(d/c))

(here one has to choose a special definition of arctan that ensures you recover the correct quadrant, and also when c = 0, define arctan(d/c) to be π/2 when d > 0, and -π/2 when d < 0 (if d = 0 as well, we just have e^0 = 1)).

the trouble is, that in the real case, we define a^x to be e^(xln(a)),

and in the complex case, log(a + bi) has some peculiarities (the complex exponential is periodic, so like trig functions, you have to restrict its domain to get a 1-1 inverse, meaining there is no single way to get a log function defined for all complex numbers, you have a choice of several log functions to choose from, called "branches").

so (a + bi)^(c + di) is not uniquely defined, you have to "choose a logarithm function" first, which usually means "choosing a domain restriction" for e^z (where z is a complex number).

even then, the logarithm function so defined has a rather undesirable property: it is discontinuous at some half-line (or ray) in the complex plane starting at the origin. for the "standard" choice, this is the negative real-axis.

it's hard to go into a lot of detail without getting too technical, but basically the problem of defining the complex power (a + bi)^(c + di) is related to the problem of assigning a unique angle to polar coordinates, if a given angle θ works, so does θ + 2πn, for any integer n.

On my keyboard, i exploit ctrl+v to get ? In complicated type, one generally makes use of i to interchange the SQRT(-a million); electric powered engineers use j (they already use i for something else). Multiply it out an same way as you could a standard monomial produced from 2 factors: ( ?15 + i)(2?15 - i) = ?15(2?15 - i) + i(2?15 - i) = 2?(15*15) - i?15 +2i?15 - i^2 ?(15*15) = 15 i^2 = -a million (definition of i) for this reason - i^2 = +a million 2i?15 - i?15 = i?15 (because 2 - a million = a million) 30 +i?15 + a million = 31 +i?15 (which will nicely be written also as) 31 + (?15)i

for (a + bi)(c + di)

FOIL it

a*c

a*di

c*bi

bi*di

the only one that is important is

bi*di

gives you

bdi^2

i^2= -1

so

bdi^2= -bd

exponetial values of i form a pattern

i^1=i

i^2= -1

i^3= -i

i^4= 1

then it repeats

i&5=i

i^6= -1

i^7= -i

i^8=1

so if you have i^53

you have to determine where ^53 falls into the pattern

53=4*13+1

i^53=i

for

(a + bi)^(c + di)

when you multiply exponents, you add them

so

(a + bi)^di * (a + bi)^c = (a + bi)^(c + di)

For multiplication, treat them normally like binomial expansion, or convert to polar form.

For exponentiation, I don't see any other method other than converting both numbers to complex form and dealing with them that way (although that is quite nasty).