(2-2i)^5

find the given power write answer in rectangular form

-128+128i

To find the given power of (2-2i)^5 and write the answer in rectangular form, we'll use the binomial theorem. The binomial theorem states that for any complex number a+bi and a positive integer n, (a+bi)^n can be expanded as:

(a+bi)^n = C(n, 0)(a^n)(b^0) + C(n, 1)(a^n-1)(b^1) + C(n, 2)(a^n-2)(b^2) + ... + C(n, n-1)(a^1)(b^n-1) + C(n, n)(a^0)(b^n)

where C(n, r) represents the binomial coefficient of n and r, which is given by the formula:

C(n, r) = n! / (r!(n-r)!)

Now, let's calculate the given power of (2-2i)^5:

In this case, a = 2 and b = -2i, and we want to find (2-2i)^5.

Using the binomial theorem, we can expand this as:

(2-2i)^5 = C(5, 0)(2^5)(-2i)^0 + C(5, 1)(2^5-1)(-2i)^1 + C(5, 2)(2^5-2)(-2i)^2 + C(5, 3)(2^5-3)(-2i)^3 + C(5, 4)(2^5-4)(-2i)^4 + C(5, 5)(2^0)(-2i)^5

Now, let's simplify each term:

C(5, 0)(2^5)(-2i)^0 = 1(32)(-2i)^0 = 32(1) = 32

C(5, 1)(2^5-1)(-2i)^1 = 5(32-1)(-2i) = 5(31)(-2i) = -310i

C(5, 2)(2^5-2)(-2i)^2 = 10(32-2)(-2i)^2 = 10(30)(-4) = -1200

C(5, 3)(2^5-3)(-2i)^3 = 10(32-3)(-2i)^3 = 10(29)(-8i) = -2320i

C(5, 4)(2^5-4)(-2i)^4 = 5(32-4)(-2i)^4 = 5(28)(16) = 2240

C(5, 5)(2^0)(-2i)^5 = 1(2^0)(-2i)^5 = 1(-2i)^5 = -32i

Now, let's add up all the simplified terms:

32 + (-310i) + (-1200) + (-2320i) + 2240 + (-32i)

Combine terms with real parts and imaginary parts separately:

Real Part: 32 + (-1200) + 2240 = 1072
Imaginary Part: (-310i) + (-2320i) + (-32i) = -2662i

Therefore, the answer to the given power (2-2i)^5 in rectangular form is 1072 - 2662i.