How do you simplify #(7+2i )/( 4+5i)#?

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Answer 1 · 2016-07-26T05:48:46

#38/41-27/41i#

You can multiply numerator and denominator by the same expression #4-5i#

#((7+2i)(4-5i))/((4+5i)(4-5i))#

#(28-35i+8i-10i^2)/(16-25i^2)#

Since #i^2=-1#, you have:

#(28-27i+10)/(16+25)#

#(38-27i)/41#

#38/41-27/41i#