What is #(4-3i)^5# in the form #a+bi# ?

1 Answer
George C.
Oct 29, 2017

#-3116+237i#

Explanation:

#(4-3i)^2 = 16-24i+9i^2 = 16-9-24i = 7-24i#

#(4-3i)^4 = (7-24i)^2 = 49-336i+576i^2 = 49-576-336i#

#= -527-336i#

#(4-3i)^5 = (4-3i)(-527-336i) = -2108-1344i+1581i+1008i^2#

#= -3116+237i#

Notes

Alternatively, consider the recursively defined pair of sequences:

#{ (a_0 = 1), (b_0 = 0), (a_(n+1) = 4a_n+3b_n), (b_(n+1) = 4b_n-3a_n) :}#

They start:

#1, color(white)(+)4, color(white)(+0)7, color(white)(0)-44, -527, -3116#

#0, -3, -24, -117, -336, color(white)(+0)237#

Then #(4-3i)^n = a_n + b_ni" "# for #n = 0, 1, 2, 3,...#

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