How do you find z such that #(4−3i) /z+(−5+2i) = (−2−2i) #?

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Answer 1 · 2016-06-19T14:18:24

#z=24/25+7/25i#

Write as:#" "" "(4-3i)/z=(-2-2i)-(-5+2i)#

#(4-3i)/z=(-2-2i)+(5-2i)#

#(4-3i)/z=(3-4i)/1#

Turn everything upside down:

#z/(4-3i)=1/(3-4i)#

#z=(4-3i)/(3-4i)#

Multiply by #1# but in the form of #1=(3+4i)/(3+4i)#, which is the complex conjugate of the denominator.

#z=(4-3i)/(3-4i)xx(3+4i)/(3+4i)=((4-3i)(3+4i))/((3-4i)(3+4i))#

Distribute:

#z=(12+16i-9i-12i^2)/(9+12i-12i-16i^2)=(12+7i-12i^2)/(9-16i^2)#

Since #i=sqrt(-1)#, we know that #i^2=-1#:

#z=(12+7i-12(-1))/(9-16(-1)) = (12+7i+12)/(9+16)=(24+7i)/25#

#z=24/25+7/25i#