How do you simplify #(7-2i)(4-i)#?

Question

1990 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-08T15:30:35

The answer is #=26-15i#

This is the multiplication of 2 complex numbers

#z_1=a_1+ib_1#

and #z_2=a_2+ib_2#

and #i^2=-1#

#z_1*z_2=(a_1+ib_1)*(a_2+ib_2)#

#=a_1a_2+ia_1b_2+ia_2b_1+i^2b_1b_2#

#=a_1a_2-b_1b_2+i(a_1b_2+a_2b_1)#

Here we have,

#(7-2i)(4-i)=28-7i-8i+2i^2#

#=28-2-15i#

#=26-15i#