- j^6 + j^ 4 - j^2
- j^8 * j^4
- (j^6 - j^ 8)/j^3
- (√(-9) + j^6) / (√(-4)
- (√(-25) + j^4) + (j^8 - √(-9))
- 4 + 8j added to 11 - 18j
- (5 - 3j) (6 -4j)
- (6 - 2j) / (5 + 3j)
- Express in polar form -6 + 8j
- Express in polar form -4 - 7j
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j^6 + j^ 4 - j^2
ReplyDeletej^2j^2j^2 + j^2j^2 - j^2
(-1)(-1)(-1) + (-1)(-1) - (-1)
(-1) + 1 - (-1)
0 - (-1)
= 1
(√(-25) + j^4) + (j^8 - √(-9))
5) (√25j + j^2j^2) + (j^2j^2j^2j^2 - √9 j)
(5j + 1) + (1 - 3j)
=( 2 + 2j)
7) (5 - 3j) (6 -4j)
30- 20j - 18j + 12j^2
30 - 38j - 12
= 42 - 38j
Q3. (j^6 - j^8)/ j^3
ReplyDeletetaking the numerator:
j^6 - j^8
(-1)(-1)(-1) - (-1)(-1)(-1)(-1)
= -1 - (+1)
= -1 - 1
= -2
taking the denominator:
= j^3
= (j^2)(j)
= (-1)(j)
= -1j
putting the expression back together:
= -2/-1j
no 6
ReplyDelete(4+8j) + (11-18j)= 15-10j
no 7
(5-3j)(6-4j) = 30-20j-18j+12j^2
= 30 - 38j -12
= 18 - 38j
no 9
ReplyDelete-6 + 8j
tan x = 8/-6
x = -53
since the phase angle is in the 2nd quadrant it will be:
180 - (-53) = 233 degrees
the manitude is:
(8^2 + -6^2)^1/2 = 10
so the complex no. in polar form will be:
10(cos 233 + j sin 233)
no 10
ReplyDelete-4-7j
the phase angle is:
tan x = opp/adj
tan x = -7/-4
x = 60
since the phase angle is in the third quadrant it will be:
x = 220 degrees
the magnitude is:
=(-7^2 + -4^2)^1/2
= (-11)^1/2
so the complex no in polar form will be:
(-11)^1/2 (cos 220 + j sin 220)
2. j^8 * j^4
ReplyDelete( j^2 * j^2 * j^2 * j^2) * (j^2 * J^2)
(-1 * -1 * -1 * -1) * (-1 * -1)
(-1) * (-1)
= 1
Q4;
ReplyDeletefirst we do the calculations inside the brackets
if we check the first bracket term
there is √(-9) which has to be made into a complex number
so √(-9)= √(-1)*√9
but √(-1) = j
therefore √(-9) = √9*j = 3j
expand j^6
which is the same as (j^2)*(j^2)*(j^2)
if j^2 = -1
then j^6 = (-1)*(-1)*(-1) = -1
the second bracket has
also √(-9) which we have shown is 3j
and j^8 is the same as j^6 x j^2
if j^6 = -1 and j^2 = -1
then J^8 = (-1) x (-1) = 1
so the question become
(3j-1)+(1+3j)
putting imaginary together and real together
we have
3j+1
3j-1
then add to give 6j
i dont undrestand this phase angle thing, can some1 pls explain in d-tails?
ReplyDelete8) (6-2j)/(5+3j)
ReplyDeleteboth terms have to be multiplied by the conjugate of the denominator..which is (5-3j)
for the numerator: (6-2j)x (5-3j)
(multiplying out the brackets)= 30-10j-18j+6j^2
= 30-28j+[6x(-1)]
= 30-6-28j
= 24-28j
for the denominator: (5+3j)x(5-3j)
[(difference of 2 squares) a^2-b^2]
= (5^2)-(3j)^2
= 25-9j^2
= 25-(9x -1)
= 25+9
= 34
then:numerator/denominator
=(24-28j)/34
taco bell for number 2 you have the right answer but the third stepis 1*1 not(-1) * (-1).
ReplyDeleteboy-b ....i think that whole phase angle thing has to do at what angle the forces start to act .........so when you find the impedence with respect to the real and imaginary numbers and plot it on the graph you find the force and the angle from the x-axis ....at least that is what i think
ReplyDelete(6) 4 + 8j +
ReplyDelete11 - 18j
= 15 - 10j
the reals are added together and th imaginary are added together
boy-b...the phase angle is easy to recognise when the graph is plotted...remember when we plot this graph and you connect the origin to the point plotted...an angle is formed between this line and the x-axis...that is usually the phase angle...it is also represented as '@' in the formula for simplifying in polar form which is.. r(cos@ + j sin@)..oh and it is usually calculated by tan@ = opp./adj. ...hope this helped alittle
ReplyDeletej^6 + j^ 4 - j^2
ReplyDeletej^6 + j^2
j^8
(j^2)(j^2)(j^2)(j^2)
(-1)(-1)(-1)(-1)
= 1
j^8 * j^4
ReplyDeletej^12
(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)
= 1
(5 - 3j) (6 -4j)
ReplyDelete30 - 20j -18j + 12j^2
30 -38j + 12(-1)
30 -38j - 12
18 - 38j
I strongly agree with Hellscream with those two questions above.
ReplyDeleteQUESTION 1
ReplyDeletej^6 + j^ 4 - j^2
just by looking at this we could see all alike variables all is j al that is need to be done is expanding into simpler and manageable form.we know j^2=-1 so we will try to get all in this form.
= (j^2)(j^2)(j^2) + j^2j^2 - j^2
= (-1)(-1)(-1) + (-1)(-1) - (-1)
= [(-1) + 1] - (-1)
= 0 - (-1)
= 1
QUESTION 2
j^8 * j^4 Using the same concept
=( j^2 * j^2 * j^2 * j^2) * (j^2 * J^2)
=(-1 * -1 * -1 * -1) * (-1 * -1)
=(-1) * (-1)
= 1
4 + 8j + 11 - 18j
ReplyDeletewhat you do is add 4 to 11 because it is real numbers then you add the imaginery which is 8j added to -18j remember look out for your positive and negative signs. They are easy to ignore when in a rush.
Question #1: j^6 + j^ 4 - j^2
ReplyDeleteRecall; j^2 = -1
which implies: j^6 = (j^2 * j^2 * j^2
= (-1 * -1 * -1)
= -1
also; j^4 = (j^2 * j^2) = (-1 * -1) = 1
with this in mind we can solve effortlessly:
(-1) + 1 - (-1) = 1
Question #9 :Express in polar form -6 + 8j
ReplyDeleteBefore attempting this question, you should plot a graph to have a visual idea of what they're asking for.
So, on the real-axis you will have 6 units left of the zero point (origin).
Also, on the imaginary-axis you will have 8 units above the zero point (origin).
Now, between the point of intersection of these two points and the origin is your magnitude.
But if you look closely, you'll see that these two points form a right-angle triangle with the magnitude.
So, using Pythagoras' Theorem we can find the hypotenuse (magnitude).
Which implies: mag = √[(-6)^2 + 8^2] = 10
Now, recall: tanθ = opp/adj
= 8/(-6)
which implies; θ = tan^-1 (8/-6)
Now, all you have to do is substitute θ and the mag in the form "r cosθ + j sinθ"
1] j^6 + j^ 4 - j^2
ReplyDeletefirst of all j^2=-1
therefore j^6=j^2 j^2 j^2
=-1 -1 -1
j^4=j^2 j^2
=-1 -1
j^2=-1
full eq= -1-1-1 + -1-1+ -1
=-1 +1- -1
0--1
=1
7] (5 - 3j) (6 -4j)
firstly expand the eq
=30-20j-18j+12j^2 j^2=-1
30+38j+12(-1)
=18+38j
people what is polar form ah doh no if u cud do ah polar form with 1 set of numbers tel me if this possible and explain
ReplyDeleteIf i'm right, polar form is just another way of expressing real and imaginery numbers. Its just in terms of sin and cos.
ReplyDeleteFirst you have to plot on a graph of imaginery against real. i.e. Real numbers an the x-axis and imaginery numbers on the y-axis.
Can someone please explain further.... I'm kinda confused!!!
1.)j^6 + j^ 4 - j^2
ReplyDelete(j^2)(j^2)(j^2)+(j^2)(j^2)-j^2
j^2=-1
(-1)(-1)(-1)+(-1)(-1)-(-1)=
-1+1+1=1
7.)(5 - 3j) (6 -4j)=
ReplyDelete=30-20j-18j+12j^2, j^2=-1
=30-38j-12
=18-38j
6.)4 + 8j added to 11 - 18j
ReplyDelete11-18j
+
4+8j
=7-10j
5 - 3j) (6 -4j)
ReplyDelete30 - 20j -18j + 12j^2
30 -38j + 12(-1)
30 -38j - 12
18 - 38j
ques1. j^6 + j^4 - j^2
ReplyDelete=j^10-j^2
=j^8
we know that j^2=(-1)
therefor (-1)(-1)(-1)(-1)= 1
Ques2.j^8 * j^4
ReplyDeletej^2=-1
(j^2)(j^2)(j^2)(j^2)*(j^2)(j^2)
(-1)(-1)(-1)(-1)*(-1)(-1)
1*1
=1
This comment has been removed by the author.
ReplyDeleteQues3. (j^6 - j^8)/ j^3
ReplyDeletej^6 - j^8
j^2=-1
(-1)(-1)(-1) - (-1)(-1)(-1)(-1)
= -1 - (+1)
= -1 - 1
= -2
since in j^3 we are only able to get one j^2 and a remainder of 1j we end up with
= j^3
= (j^2)(j)
= (-1)(j)
= -1j
since the j^6 - j^8 is divided by j^3 we put the -2 over the -1j
= -2/-1j
Ques.7(5 - 3j) (6 -4j)
ReplyDeletewe open out the brackets first i think
(5-3j)(6-4j)
30-18j-20j+12j^2
12j^2-38j+30
12(-1)-38j+30
-12+30-38j
18-38j
Ques.8(6 - 2j) / (5 + 3j)
ReplyDelete(6-2j)x (5-3j)
30-10j-18j+6j^2
30-28j+6x(-1)
30-6-28j
24-28j
(5+3j)x(5-3j)
25+15j-15j-9j^2
25-9j^2
25-(9x -1)
25+9
34
=(24-28j)/34
for question # 3,
ReplyDeleteuse j^2 = -1 and √-1 = j
When cmplex numbers are given with brackets, expand and then simplify....
ReplyDeleteAlways remeber ther j^2 = -1
Square root j = -1/2
1.Express in polar form -6 + 8j
ReplyDeleteTan θ = opp
___
Adj
Tan θ = 8
__
-6
θ tan-1 = 8
__
-6
Θ = 53.13
√100 (cos 53.13 + j sin 53.13)
Express in polar form -4 - 7j
Tan θ = opp
___
Adj
Tan θ = -7
__
-4
θ tan-1 = -7
___
-4
Θ = 60.26
√114 (cos 60.26 + j sin 60.56)
need help with this question
ReplyDelete25-14j/4+3j
This comment has been removed by the author.
ReplyDelete1. j^6 + j^4 - j^2
ReplyDelete=(j^2 x j^2 x j^2) + (j^2 x j^2) - j^2
=-1 + 1 + 1
=1
3. j^6 - j^8 / j^3
ReplyDelete= (j^2 x j^2 x j^2) - (j^2 x j^2 x j^2 x j^2) / j^2 x j
=-1 - 1 / -1j
=-2 / -1j
4. √-9 + j^6 / √-4
ReplyDeleteSince √-1 =j
= √-1 x 9 + (j^2 x j^2 x j^2) / √-1 x 4
=√-9j -1 / √4j
=3j - 1 / 2j
7. (5-3j)(6-4j)
ReplyDelete=30 -20j -18j +12j^2
=30 -38j +12j^2
=30 -38j (+12 x -1)
=30 - 28j -12
=18 -38j
8. (6 -2j) / (5 +3j)
ReplyDelete=(6 -2j / 5 +3j) x (5 -3j / 5 -3j)
={(6 -2j) x (5-3j)} / {(5 +3j) x (5 -3j)}
=30 -18j -10j +6j^2 / 25 -15j +15j -9j^2
=30 =28j +6j^2 / 25 -9j^2
=30 -28j -6 / 25 +9
=24 -28j / 34