How do you simplify #-9-9i div -1-8i#?

2 Answers
Apr 15, 2016

Answer:

#-9-9idiv-1-8i= color(green)(81/65-63/65i)#

Explanation:

#-9-9i div -1-8i = (-9-9i)/(-1-8i)#

#color(white)("XXX")=(-9-9i)/(-1-8i)*(-1+8i)/(-1+8i)#

#color(white)("XXX")(9+9i-72i+72)/(1+64)#

#color(white)("XXX")=(81-63i)/65#

Apr 15, 2016

Answer:

#−9−9i div −1−8i=81/65-63/65i#

Explanation:

#−9−9i div −1−8i# can be written as #(−9−9i)/(−1−8i)#

To simplify we need to multiply numerator and denominator by the complex conjugate of the denominator.

Complex conjugate of a number #a+bi# is #a-bi#,

hence in above case one needs to multiply by #-1+8i#

Hence #(−9−9i)/(−1−8i)#

= #((−9−9i)xx(-1+8i))/((−1−8i)xx(-1+8i))#

= #(9-72i+9i-72i^2)/((-1)^2-(8i)^2)#

= #(9-72i+9i-72(-1))/(1-64(-1))#

= #(9-72i+9i+72)/(1+64)#

= #(81-63i)/65#

= #81/65-63/65i#