How do you solve #z( 3 + 2i) = (2 — 3i)#?

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Answer 1 · 2015-12-29T16:40:56

#z= i#

#z(3+2i) = (2-3i)#

To solve for #z# divide both sides by #3+2i# like this

#(z(3+2i))/(3+2i) = (2-3i)/(3+2i)#

#=>z = (2-3i)/(3+2i)#

Then multiply by the conjugate of the denominator to simplify the expression

#z= ((2-3i)/(3+2i))*color(red)(((3-2i)/(3-2i))#

Then multiply and combine like terms like this

#z= (6-4i-9i+6i^2)/(9-6i+6i-4i^2)#

#=> (6 - 13i - +6i^2)/(9-4i^2)#

Replace #i^2 = -1# then simplify

#=>z = (6 -13i +6*color(blue)(-1))/(9-4*color(blue)(-1))#

#=>z = (6-6-13i)/(9+4)#

#=>z = (-13 i)/(13) = i#

#z= i#

I hope this helps :)