find the argument of this
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(iii) (1+i)(3+i) =3+i+3i+i^2 =3+4i-1 (i^2 =-1 ) =2+4i= 2√5(1/√5 +2/√5i) => tanA=2/√5/(1/√5)=2 .
(iv) √3+i/-(1+i√3) = (√3+i)(1-i√3)/-(1+i√3)(1-i√3) = √3-3i+i-i^2(3) =√3-2i+3 /-(1-i^2(9))=3+√3-2i/-(1+9) =-(3+√3-2i)/10 => tanA= -2/3+√3 . Hope it helps you....
(iv) √3+i/-(1+i√3) = (√3+i)(1-i√3)/-(1+i√3)(1-i√3) = √3-3i+i-i^2(3) =√3-2i+3 /-(1-i^2(9))=3+√3-2i/-(1+9) =-(3+√3-2i)/10 => tanA= -2/3+√3 . Hope it helps you....
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