How do you divide #(5i)/(6+8i)#?

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Answer 1 · 2016-12-10T11:18:11

The answer is #=0.4+0.3i#

#z_1/z_2#

You multiply the numerator and denominator by the conjugate of the denominator

#(z_1*barz_2)/(z_2*barz_2)#

If #z=a+ib#

Then, #barz=a-ib#

and #i^2=-1#

#(a+b)(a-b)=a^2-b^2#

Here,

#(5i)/(6+8i)=((5i)(6-8i))/((6+8i)(6-8i))#

#=(30i-40i^2)/(36-64i^2)#

#=(40+30i)/100#

#=0.4+0.3i#