How do you simplify #(4+ 2i) /( -1 + i)#?

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Answer 1 · 2018-05-02T22:48:17

#(4+2i)/(-1+i) | *(-1-i)#

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

#(-2i^2-6i-4)/(1-i^2)#

#(2-6i-4)/(1+1)#

#(-2-6i)/(2)#

#=-1-3i#

We want to get rid of #i# in the the bottom of the fraction in order to get it on Certesian form. We can do this by multiplying with #(-1-i)#.

This will give us,

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

#(-2i^2-6i-4)/(1-i^2)#

Out from here we know that #i^2=-1# and #-i^2=1#. So we can get rid of the #i^2# too. Leaving us to

#(-2-6i)/(2)#

#=-1-3i#