Question

1. Simplify the following expression a. j22 = b. 2j? +3j* = C. (5j?) = d. j j = f. 3j(2j? –5j)= g. 6xjº +2yj“ + 4xj +5yje =

Answer 1

233 by 4, t ind reminder step -0: Divide 133 ID 1270935 24014 4) 233 (58 20 33 2706 (34)59 - j' O 270 step@: 5233 becomes eqrel to 2706 20 233 6) 270 270 since 27 58 :- 27 :233 %3D Ans 27 29 252 + 3j3 = 27 divide power of i by 4. Tay step O: 2. to 2- as it is in locoest can't Here . already. form 2. 3. %3D 2j +3) step © put j? = -1 j= jj=-j value and (1,012 2j*+3j* 2j +2j =-2-3j 2(-1) + 3(-j) Ans 2020-1-16 04:50

step O: Enpand (5j) to oder onginal form . (5j) = 5x j = 125 j? step@: 4s know : we . put j%= -j = 125 (-j) 125j => -> 5i) = n5(-i) = 125 ALs know: stepD: As am * an = j(G+3) it power of j Te. 10 by 4 t find s tepe: Dinde neminder 4J 19 (2 2. Thus, jio lj)"j". j? step@ =-> der j" = = -1 =j? = (-1) >j7. j = - 2020-1-16 04:50

step0: Dinide power o j by 4t find neminder 4)3 (0 step O: 16 stepO -> = om-n step@: Ac 4-3 (d. 16 -> Ans stepC 3j(2j²-5i)= step O: a"x0 = dm+n) As know; stepe : 3j'. (2j-5j) = 3j *2j - 1j*5j = 6 je-(3¥5) fit1) 6 j3 - 15 jz stet power ey i by 4 t find reminder "):(0 step@. Dinde =-> 4) 2 (0 6j- 15j? = 16(i) - 15(3?) 2020-1-16 04:51

Stels ® : know: j2=-1 j3 = j'j = -j and 6-1sj 6 (-5)- 15(-1) = 6j+15 3i(2j-5j) =-6j +15 %3D Ans ) 6xj+1" + 4xi + 5yj= step0: since power of j is in lowest form so jut ut value that we know that: j'-j: j=t; j² =-j 6 x j3 = 6x (-j) = -6x j = 2y(1) = 24 24lj") => 24j" = %3D > 5yj)" 5y (1) = 5y abome equation step : put all : 6x j% + ry j"+uxj+5yj8= -62j +2y+uxj values in +5y - 2 1 J+7y Ionj'tayj" ynj+syj8 = ay-2xj} Ay 2020-1-16 04:51