How do you simplify #z^ { 3} - 2( 2+ i ) z ^ { 2} + ( 5- 8i ) z - 10i = 0#?

1 Answer
Tamir E.
Nov 16, 2017

#a^3-3ab^2-4a^2+4b^2+5a+8b=0#
#3a^2b-b^3-12ab-2a^2+2b^2+5b-8a-10=0#

Explanation:

#z=(a+bi)#
#(a+b)^3=a^3+3a^2b+3ab^2+b^3#

#z^3=(a+bi)^3=a^3+3a^2bi+3a(bi)^2+(bi)^3#
#=> z^3=a^3+3a^2bi-3ab^2-b^3i#
#=> z^3=a^3-3ab^2+3a^2bi-b^3i#
#=> z^3=(a^3-3ab^2)+(3a^2b-b^3)i#

#z^2=(a+bi)^2=a^2+2abi-b^2#

#-2(2+i)=-4-2i#

#(-4-2i)(a^2+2abi-b^2)=-4a^2-8abi+4b^2-2a^2i-4abi+2b^2i#
#= -4a^2-12abi+4b^2-2a^2i+2b^2i#
#= -4a^2+4b^2-12abi-2a^2i+2b^2i#
#= (-4a^2+4b^2)+(-12ab-2a^2+2b^2)i#

#(5-8i)(a+bi)=5a+5bi-8ai+8b#
#=5a+8b+5bi-8ai#
#=(5a+8b)+(5b-8a)i#

So we know now:
#z^3-2(2+i)z^2+(5-8i)z-10i=0#
#=> (a^3-3ab^2)+(3a^2b-b^3)i+ (-4a^2+4b^2)+(-12ab-2a^2+2b^2)i+(5a+8b)+(5b-8a)i-10i=0#
#=> (a^3-3ab^2)+ (-4a^2+4b^2)+(5a+8b)+(3a^2b-b^3)i+(-12ab-2a^2+2b^2)i+(5b-8a)i-10i=0#
#=> (a^3-3ab^2-4a^2+4b^2+5a+8b)+(3a^2b-b^3-12ab-2a^2+2b^2+5b-8a-10)i=0#

Now we solve:
all WITHOUT i = 0
all WITH i = 0
#=>#
#(a^3-3ab^2-4a^2+4b^2+5a+8b=0)#
AND
#(3a^2b-b^3-12ab-2a^2+2b^2+5b-8a-10=0)#

#a^3-3ab^2-4a^2+4b^2+5a+8b=0#
#3a^2b-b^3-12ab-2a^2+2b^2+5b-8a-10=0#

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