can someone please explain how to find the power of a complex number?

Question

z=(3+i)^3
I can't find any examples of this variety of problem in this format, where the second number is an irrational/complex number. Can anyone explain the steps in solving a problem like this please?

Answer 1

multiply the number as many times as the power indicates:
(3 + i)^3 = 27 + 3(9)i + 3 (3) (-1) + i^3
= 27 -9 + 27i - i
= 18 + 26i

Answer 2

howdy mate, via definition f(z) = e^(z) = sum ( z^(n) / n ! , n = 0 to infinity) f ' (z) = d/dz [ e^(z) ] = d/dz [ sum ( z^(n) / n ! , n = 0 to infinity) ] f ' (z) = d/dz [ e^(z) ] = sum ( d/dz [ z^(n) / n ! ] , n = 0 to infinity) i visit state here without evidence d/dz [ z^(n) / n ! ] = n * z ^ (n - a million) / n ! consequently, f ' (z) = d/dz [ e^(z) ] = sum ( n * z ^ (n - a million) / n ! , n = 0 to infinity) be conscious on the term n = 0 ; n * z ^ (n - a million) / n ! = 0, consequently isn't required interior the sum, subsequently, f ' (z) = d/dz [ e^(z) ] = sum ( n * z ^ (n - a million) / n ! , n = a million to infinity) be conscious n / n ! = n / ( n * (n - a million) ! ) = a million / (n - a million)! consequently, f ' (z) = d/dz [ e^(z) ] = sum (z ^ (n - a million) / (n - a million) ! , n = a million to infinity) substitute the index from n to m, wherein m = n - a million, consequently at n = a million , m = 0 ; as n --> infinity m --> infinity consequently, f ' (z) = d/dz [ e^(z) ] = sum (z ^ (m) / m ! , m = 0 to infinity) which you will needless to say see is the definition of e ^ (z) consequently, f ' (z) = d/dz [ e^(z) ] = e^(z) desire this permits, David playstation - with a view to verify that this by-product is defined for all values of z being a complicated variety the least confusing technique of attack is to hire the Cauchy - Riemann eqns, please message me in case you require any help with this.

Answer 3

You start by expanding it just like you would (x + 2)^3
a.k.a. binomial expansion.

(3 + i)^3 = 1*3^3 + 3*3^2i + 3*3i^2 + 1*i^3

i = √(-1) so i^2 = -1
i^3 = i * i^2 = -i
Use these to substitute for i^2 and i^3. combine real and imaginary parts and you have your answer.