How do you divide #-2-2i div 5-2i#?

1 Answer
Feb 1, 2016

Answer:

# -6/29 - 14/29 i#

Explanation:

When dividing complex numbers ,multiply numerator and

denominator by the complex conjugate of the denominator.

If a + bi is a complex number then#color(red)(" a - bi is conjugate")#

This ensures the denominator is real.

as # (a+bi)(a-bi) = a^2 + b^2 color(black)( " which is real")#

[Note: # i^2 =( sqrt-1 )^2 = -1 ]#

the conjugate of 5 - 2i is 5 + 2i .

# rArr( (-2-2i)(5+2i))/((5-2i)(5+2i)) #

( distribute to obtain) #(-10-14i-4i^2)/(25-4i^2) #

# =( -6-14i)/29 = -6/29 -14/29 i#