<PRE><B>Question 213435</B>
Divide: {{{(4+2i)/ (4-2i)}}}


Step 1.  Multiply by 1 where {{{(4+2i)/(4+2i)=1}}} so that the denominator is not complex.


Step 2.  This will result in the following steps


{{{(4+2i)/(4-2i)=((4+2i)/(4-2i))*(4+2i)/(4+2i)}}}


Now the denominator will be


{{{(4-2i)(4+2i)=(4)(4)-(4)(2i)+(2i)(4)-(2i)(2i)}}}


{{{(4-2i)(4+2i)=16-(4)i^2=16-4*(-1))=20}}} where {{{i=sqrt(-1)}}}


{{{(4+2i)/(4-2i)=(4+2i)*(4+2i)/20}}}


Now the numerator will be


{{{(4 + 2i)^2=16+2i+2i+4i^2=16-4+4i=12+4i}}}


{{{(4+2i)/(4-2i)=(12+4i)/20=(3+i)/5}}}



I hope the above steps were helpful.


For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.


Good luck in your studies!


Respectfully,
Dr J

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