#z# is a complex number, #z=a+ib# and #(2+i)z=1-z#, what is #z# in terms of #i# ?

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Answer 1 · 2016-03-09T22:47:17

#z = 3/10-1/10i#

#(2+i)z = 1-z#

#=> (2+i)z + z = 1#

#=> (2+i+1)z = 1#

#=> (3+i)z = 1#

#=> z = 1/(3+i)#

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

#=(3-i)/10#

#=3/10-1/10i#

#:. z = 3/10-1/10i#