How do you divide #(5-8i )/( 8+5i)#?

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Answer 1 · 2016-01-28T09:52:00

#(5-8i)/(8+5i)=-i#

Given:

#z=z_1/z_2# with #z_1,z_2 in CC#

you can compute the Division of Complex Numbers multiplying both denominator an numerator by #bar(z_2)#, the complex conjugate of #z_2#

If #z_2=a+ib => bar(z_2)=acolor(red)-ib#

#:.(5-8i)/(8+5i)=(5-8i)/(8+5i)times(8-5i)/(8-5i)=#

#=(5*8-5*5i-8i*8+8i*5i)/(8^2-(5i)^2)=(40-25i-64i+40i^2)/(64-25i^2)#

remembering that #i^2=-1#

#=(cancel40-cancel40-89i)/(64+25)=-(cancel(89)i)/cancel(89)=-i#