find the modulus Z=(2-i)(5+12i)/(1+2i)³
in order of operations by Level 1 User (180 points)

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1 Answer

(2-i)(5+12i)=10+19i+12=22+19i.

(1+2i)^3=1+6i+3(2i)^2+(2i)^3=1+6i-12-8i=-11-2i=-(11+2i); -1/(11+2i)=-(11-2i)/(121+4)=-(11-2i)/125.

Z=-(22+19i)(11-2i)/125=-(242+165i+38)/125=-(280+165i)/125=-(56+33i)/25.

|Z|=√((-56)^2+(-33)^2)/25=√4225/25=65/25=13/5.
by Top Rated User (1.1m points)

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