Find the modulus and argument of z = ( 2- i)(1-3i) and (-1+2i)(1+i) all over (-2-3i).

The question above is all complex numbers which features finding the modulus and arguments.
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1 Answer

z=((2-i)(1-3i)+(-1+2i)(1+i))/(-2-3i)=

-(2-3-7i-1-2+i)/(2+3i)=-(-4-6i)/(2+3i)=

2(2+3i)/(2+3i)=2.

Therefore |z|=2 and arg(z)=0

by Top Rated User (1.2m points)

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