How do you simplify #(4+ 3i) div (2 - i)#?

1 Answer
Jul 13, 2016

Answer:

#(4+3i)/(2-i)=1+2i#

Explanation:

The complex conjugate of a complex number #a+bi# is denoted #bar(a+bi)# and is given by #bar(a+bi) = a-bi#. A useful property of the complex conjugate is that for any complex number #z#, we have #zbar(z) in RR#. We will use this property to simplify the given expression by multiplying the numerator and denominator by the conjugate of the denominator.

#(4+3i)/(2-i) = ((4+3i)(2+i))/((2-i)(2+i))#

#=(8+6i+4i-3)/(4+2i-2i+1)#

#=(5+10i)/5#

#=1+2i#