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 $i^2=-1$

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

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