How do you simplify #\frac { 14+ 8i } { 2- i }#?

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Answer 1 · 2017-06-07T08:39:47

#4+6i#

Remember that #i^2=-1#

#(14+8i)/(2-i)#

#(14+8i)/(2-i)*(2+i)/(2+i)#

#(28+30i+8i^2)/(4-i^2)#

#(28+30i+8(-1))/(4-(-1))#

#(20+30i)/5#

#4+6i#