Question

Please answer the giver question its very urgent .....plzzzz I beg you





Reduce ( 1 / 1-2i + 3 / 1+i) ( 3+4i / 2- 4i ) to standard form ..... here i refers to iota

Answer 1

SOLUTION

=) [(1+i)+3(1-2i)/(1-2i)(1+i)]×(3+4i/2-4i)

{by taking L.C.M}

=)[(1+i)+(3-6i)/1+i-2i-2i^2][3+4i/2-4i]

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

=)[4-5i/1-i+2][3+4i/2-4i]

=)[4-5i/3-i][3+4i/2-4i]

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

=)12+16i-15i-20i^2/6-12i-2i+4i^2

=)12+i-20(-1)/6-14i+4(-1)

=)12+i+20/6-14i-4= 32+1/2-14i

On rationalising, we get

=)32+i/2-14i×2+14i/2+14i

=)32(2+14i)+i(2+14i)/2(2+14i)-14i(2+14i)

=) 64+448i+2i+14i^2/4+28i-28i-196i^2

=)64+450i+14(-1)/4-196i^2

=)64-14+450i/4-196(-1)

=)50+450i/4+196

=) 50+450i/200

hope it helps ☺️

Answer 2

Answer:

don't worry I will do ...........