(2-2i)^5 What is this in standard notation?

1 Answer
Steve M
Apr 5, 2018

# (2-2i)^5 = -128+128i #

Explanation:

We seek:

# (2-2i)^5 #

We can expand using the Binomial Theorem:

# (2-2i)^5 = (2)^5(-2i)^0 + 5(2)^4(-2i)^1 + #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 10(2)^3(-2i)^2 + 10(2)^2(-2i)^3 + #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5(2)^1(-2i)^4 + (2)^0(-2i)^5 + #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 32 + 5(16)(-2i) + 10(8)(4i^2) + #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 10(4)(-8i^3) + 5(2)(16i^4) + (-32i^5)#

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 32 - 160i + 320(-1) + #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (-320)(-i) + (160)(1) + (-32)(i) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 32 - 160i - 320 + 320i + 160 -32i #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ = -128+128i#

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