SOLUTION: hai i am doing B.Sc(maths) Ist year at madurai kamaraj university in correspondence. this question is from trigonometry, complex numbers find the modulus and amplidude form

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Question 56685: hai
i am doing B.Sc(maths) Ist year at madurai kamaraj university in correspondence.
this question is from trigonometry, complex numbers
find the modulus and amplidude form of (1-i)(1-2i)/(1+3i)
i have tried this one
(1-i)(1-2i)/(1+3i) = (1-i)(1-2i)(1+3i)/(1-3i)(1+3i)(multiplyingnumerator & dinominator by the conjugate)
=(8+6i^3)/10
(1-i)(1-2i)/(1+3i) =(4-3i)/5
r=square root(x^2+y^2)
=square root ((4/5)^2+(-3/5)^2)
r=square root (1)
cos theta=x/r
=4/5 square root (1)
sin theta=y/r
=-3/5 square root(1)

Answer by venugopalramana(3286) (Show Source):
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hai
i am doing B.Sc(maths) Ist year at madurai kamaraj university in correspondence.
this question is from trigonometry, complex numbers
find the modulus and amplidude form of (1-i)(1-2i)/(1+3i)
i have tried this one
(1-i)(1-2i)/(1+3i) = (1-i)(1-2i)(1+3i)/(1-3i)(1+3i)(multiplyingnumerator & dinominator by the conjugate) ......GOOD
=(8+6i^3)/10.........................VERY GOOD
(1-i)(1-2i)/(1+3i) =(4-3i)/5 ..OK..BUT SIMPLER IS 8+6I^3=8+6I*I^2=8-6I=2(4-3I)
r=square root(x^2+y^2)...............GOOD
=square root ((4/5)^2+(-3/5)^2).................GOOD
r=square root (1) =OK.....PUT IT AS = 1
cos theta=x/r...NO WHY?
PUT TAN THETA = -3/4 AND THETA = ARC TAN(-3/4) IS THE AMPLITUDE
WELL ONE KEEP IT UP..YOU NEED JUST A LITTLE BRUSHING UP..
=4/5 square root (1)
sin theta=y/r
=-3/5 square root(1)