Z1= 8-3 j , Z2 = 7+5j , Z3 = -9-9j?

Question

i)Find 8 Z1 + 2 Z2 - 2 Z3
ii)Find Z1 Z2
iii)Find Z2/Z3
iv)Find (2Z1 - 2Z3)/Z2)
(V)Find (8Z1 - 2Z2)/(2Z2 + 2Z3)

Answer 1

Z1= 8-3 j , Z2 = 7+5j , Z3 = -9-9j?

i)Find 8 Z1 + 2 Z2 - 2 Z3
8(8-3j)+2(7+5j)-2(-9-9j)
=64-24j+14+10j+18+18j
=96+4j

ii)Find Z1 Z2
(8-3j)(7+5j)
=56-21j+40j-15j^2
=71+19j

iii)Find Z2/Z3
(7+5j)(9-9j)/162
=63+45j-63j+45/162
=(108/162)-(18/162)j

iv)Find (2Z1 - 2Z3)/Z2)
16-6j-18-18j/(7+5j)
=(-2-24j)(7-5j)/74
=-12-168j+10j-120/74
=(-132/74)-(158/74)j

(V)Find (8Z1 - 2Z2)/(2Z2 + 2Z3)
64-24j-14-10j/14+10j+18+18j
=50-34j/32+28j
=50-34j/4(8+7j)
=(50-34j)(8-7j)/452
=400-272j-350j-238/452
=(162/462)-(622/462)j

check the calculations for typos.i don't use calculators

Answer 2

Wouldn't you just substitute? So for (i) you'd have:

8 Z1 + 2 Z2 - 2 Z3 =
8(8-3j) + 2(7+5j) - 2(-9-9j) =
64 - 24j +14 +10j +18 +18j =
64 +14 +18 - 24j +10j +18j =
96 + 4j

Solve (ii) through (v) similarly -- just substitute, then plug-n-chug.

Hope that helped!

~ ♥ ~

Answer 3

4

Answer 4

Do you mean how many different numbers you can make that include ALL these digits? If so, then the number of possibilities equals the number of possibilities for the first number multiplied by the number of possibilities for the second number etc up until the last digit. Assuming we can't use zero as the first digit: 9 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 9,000,000,000

Answer 5

just plug and chug

i) 8(8-3j) + 2(7+5j) - 2(-9-9j)=
64-24j+14+10j+18+18j=
=96+4j

ect


it looks like this is a vector problem, and since all are in terms of j, then they can just be added, divided.... whatever

Answer 6

i). 8(8-3j) + 2(7+5j) - 2(-9-9j)
= 64 - 24j + 14 + 10j + 18 + 18j
= 4j + 96

ii). (8-3j)(7+5j)
= 56 + 40j - 21j - 15j^2
= -15j^2 + 19j + 56

Answer 7

im a year 8 student-i did this in school only times and divide-to show the simplist form.
18-18j
i)96+4j
ii)71+19j
iii)-2/3+1/9j
iv)149-43j)/37
v)(9+67j)/10

i think its right.and i didnt use a calculater.

Answer 8

if Z1 is complex number and is equals to a+bj and Z2 is same as Z1 and equal to c+dj we will have:
Z1+Z2=(a+c)+(b+d)j
Z1-Z2=(a-c)+(b-d)j
because of j*j=-1 => Z1*Z2=(a*c-b*d)+(a*d+c*b)j

for calculate of Z1/Z2 we will multiply Z1 and Z2 in conjugate of Z2 (because denominator be a real number).

conjugate of Z2 is c-dj

Answer 9

For addition or subtraction, add the correct parts together (add/subtract all imaginary parts, add/subtract all real parts).

For multiplication, use FOIL, and you need to remember that j^2=-1

For division, multiply the numerator and denominator by the conjugate of the denominator (if denominator is (a + bj), then multiply numerator and denominator by (a - bj))


To multiply by a number, simply multiply the real and imaginary parts by that number.

Answer 10

8Z1=64-24j
2Z2=14+10j
2Z3= -18-18j

i)8Z1+2Z2-2Z3
=64-24j+14+10j+18+18j
=96+4j

ii)Z1*Z2=(8-3j)(7+5j)
=56+19j-15j^2
=71+19j

iii)Z2/Z3
=(7+5j)/(-9-9j)
=(7+5j)(-9+9j)/(-9-9j)(-9+9j)
=(-63+18j+45j^2)/(81-81j^2)
= (-108+18j)/162
= -2/3+1/9j

iv)2Z1-2Z3=2(8-3j+9+9j)
=(34+12j)
(2Z1 - 2Z3)/Z2)
=(34+12j)/(7+5j)
=(34+12j)(7-5j)/(7+5j)(7-5j)
=(238-86j-60j^2)/(49-25j^2
=(298-86j)/74=(149-43j)/37

(v)(8Z1-2Z2)
=64-24j-(14+10j)
=50-34j
(2Z2+2Z3)=14+10j-18-18j
= -4 -8j
(8Z1 - 2Z2)/(2Z2 + 2Z3)
=(50-34j)/(-4-8j)
=(50-34j)(-4+8j)/(-4-8j)(-4+8j)
=(72+536j)/80=(9+67j)/10

i hope that this helps