Evaluate
- j^5 + j^ 3 - j^7
- j^5 * j^3
- (j^5 - j^ 7)/j^3
- (√(-9) + j^5) / (√(-4)
- (√(-25) + j^3) + (j^7 - √(-9))
- 4 + 7j added to 11 - 21j
- (5 - 9j) (6 -3j)
- (6 - 7j) / (4 - 3j)
- Express in polar form 10 + 18j
- Express in polar form -14 - 7j
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9) Express in polar form 10 + 18j
ReplyDeletetan ang= opp/adj
= 18/10
tan^-1 18/10
= 8.68 deg
hyp= √424
therefore its √424( cos 8.68+ jsin 8.68)
how is the unknown angle for tan 1?
ReplyDeleteQ1. j^5 + j^3 - j^7
ReplyDeleteusing the fundamentals: j^2=-1
j^5 is broken down into the amount of j^2 that can go into the j^5 and the sama applies to the other imaginery numbers
j^5= (j^2)(j^2)(j)
= (-1)x(-1)x(j)
= 1j
j^3= (j^2)x(j)
(-1)x(j)
= -1j
j^7= (j^2)(j^2)(j^2)(j)
= (-1)x(-1)x(-1)x(j)
= -1j
putting the numbers back together now:
1j + (-1j) - (-1j)
= 1j -1j + 1j
= 1j
question 4: (√(-9) + j^5) / (√(-4)
ReplyDeleteremember that j= √-1 and j^2= -1
look for notted terms and sheds.
simplify the √ of the negative number
√(-9)= √(-1)*√(9)= 3j
√(-4)= √(-1)*√(4)= 2j
simplify j^5 to give the number of possible j^2
j^5 = j^2*j^2*j
= (-1)*(-1)*j
= 1j or j
rewrite the equation to give:
3j + j / 2j
= 4j/2j
= 2
or it could have been done by using the conjugate to make the denominator a real number
ReplyDelete0 + 4j 0 - 2j
------------- * ----------
0 + 2j 0 - 2j
denominator
remember (a+b)(a-b)= a^2 - b^2
so: (0+2j) (0-2j)= 0^2 - (2j)^2
= 0 - 4j^2
= 0 + 4
= 4
numerator
(0+4j)*(0-2j)
expand the brackets
= 0 - 0 + 0 - 8j^2
= 8
therefore the solution would be
numerator/ denominator
= 8 / 2
= 4
the equation should be
ReplyDelete0 + 4j * 0 - 2j
________ ______
0 + 2j 0 - 2j
(4 + 7j) + (11 - 21j)
ReplyDelete15 + -14j
(5 - 9j) (6 -3j)
ReplyDelete30-15j-54+27j^2
30-69j-27
= 3-69j
(6 - 7j) / (4 - 3j)
ReplyDeleteur conjugate would be (4 +3j)/(4 - 3j)
the dem 1st.. so it (4 - 3j)(4 +3j)
= 4^2 + 9 = 13
the multiply (6 - 7j) (4 +3j)
= 24+18j-28j=21j^2
=24-10j+ 21
=45 + 21j/ 13
no 1
ReplyDeletej^5 + j^3 - j^7
we have to work out addition first:
= j^8 - j^7
= j
and we know that j is:
= (-1)^1/2
no 5
ReplyDeletethe first thing to do is to work out anything that that can be worked out. in other words remove the shed!!
(5j + j^3) + (j^7 - 3j)
= 2j + j^10
= -1 +2j
2. j^5 * j^3
ReplyDelete( j^2 * j^2 * j) * (j^2 * J)
(-1 * -1 * j) * (-1 * j)
(1*j) * (-1*j)
= 2j^2
= 2 (-1)
= -2
no 10
ReplyDelete(-14-7j)
the phase angle is:
tan x = opp/adj
tan x = -7/-14
x = 27
since th phase angle is in the third quadrant it will be:
x = 180 + 27 = 207 degrees
the magnitude is:
=(-14^2 + -7^2)^1/2
= (-21)^1/2
so the complex no in polar form is:
(-21)^1/2 (cos 207 + j sin 207)
from what misticwings* said bout the comment above
ReplyDeletehow could the -21^1/2 could be changed into a real/imaginary change into the polar form
sum1 plz explain
to write thins in polar form 10 + 18j
ReplyDeletepolar form is r(cos x+ j sin x)
the first step is to find r that is the magnitude
using Pythagoras theorem
10^2+ 18^2=424
the square root of 424
r=20.6
the next step is to find the angle x
tan x = opposite / adjacent
tan x = 18/10
x = tan-1 1.8
x = 60.9 degrees
answer in polar form is
20.6(cos 60.9+ j sin 60.9)
3) (j^5-j^7)/j^3
ReplyDeletethis can be broken down to:
[(j^2 x j^2 x j)-(j^2 x j^2 x j^2 x j)]/j^2 x j
(substitute j^2=-1):
=[((-1)x(-1)x j) -((-1)x(-1)x(-1)x j)]/((-1) xj)
(multiply out the brackets):
= (j- (-j))/(-j)
= (j+j)/(-j)
= 2j/(-j)
(dividing 2j by -j cancels out the j's resulting in..)
= -2
7.(5 - 9j) (6 -3j)
ReplyDelete=30-15j-54j+27j^2
j^2=-1
=30-27-15j-54j
=3-69j
j^5 + j^ 3 - j^7
ReplyDeletej^8 - j^7
= j
j^5 * j^3
ReplyDeletej^8
(j^2)(j^2)(j^2)(j^2)
(-1)(-1)(-1)(-1)
= 1
4 + 7j added to 11 - 21j
ReplyDelete4 + 7j + 11 - 21j
15 -14j
B) j^5 * j^3
ReplyDeletefrom prior knowledge we know that when we multiply two like terms the powers can be added, therefore it will equal to:
j^8
and j^2 = -1
therefore
(-1)(-1)(-1)(-1)
=1
In polar form 10 + 18j
ReplyDeletetan (ang) = opp/adj
= 18/10
(ang) = tan^-1 18/10
= 60.95
hyp^2 = opp^2 + adj^2
= 18^2 + 10^2
hyp = √424
= 20.59
therefore its 20.59(cos 60.95 + jsin 60.95)
4 + 7j + 11 - 21j
ReplyDelete15 -14j
Express in polar form 10 + 18j
ReplyDeleter = root of 10^2 + 18^2 = root 424
tan teta = 18/10= 60.9 degrees
in polar for it would be
√424 ( cos 60.9 degrees + j sin 60.9 degrees)
(5 - 9j) (6 -3j)
ReplyDelete5 * 6 + 5 * -3j -9j * 6 -9j * -3j
The result of the answer above which darky haven't yet completed is
ReplyDelete= 30 - 15j - 54j + 27j squared
= 30 - 69j + 27(-1)
= 3 - 69j
4.)(√(-9) + j^5) / (√(-4)
ReplyDelete√(-9)= √(-1)*√(9)= 3j
√(-4)= √(-1)*√(4)= 2j
j^5 = j^2*j^2*j
= (-1)*(-1)*j
= 1j or j
rewrite the equation to give:
3j + j / 2j
= 4j/2j
= 2
1.)j^5 + j^ 3 - j^7=
ReplyDelete(J^2)(J^2)(J)+(J^2)(J)-(j^2)(J^2)(J^2(J).j^2=-1
=(-1)(-1)(J)+(-1)(J)-(-1)(-1)(-1)(J)
=1J-1J+1J
=1J
CAN SOMEONE SHOW ME HOW TO DO Q5
ReplyDelete#8
ReplyDelete(6 - 7j)/(4 - 3j)
u multiply it by the conjugate
(6 - 7j)/(4 - 3j)]* [(4 + 3j)/(4 + 3j)]
simplify the denominator
4^2 - 3j^2 {j^2 = -1}
=16 - 3(-1) {16 + 3}
=19
now simplify the numerator:
(6 - 7j)*(4 + 3j)
= (24 - 28j + 18j - 21j^2) {j^2 = -1}
= 24 - 10j - 21(-1)
= 24 - 10j +21
= 45 - 10j
now, divide the numerator by the denominator
=1/19*(45 - 10j)
9) Express in polar form 10 + 18j
ReplyDeletehyp = r √Real+ imaginery
therefore its √r( cos angle + j sin angle)
Set of 10
ReplyDeleteExpress in polar form 10 + 18j
Tan θ = opp
___
Adj
Tan θ = 18
___
10
θ tan-1 = 18
___
10
Θ = 60.95
√424 (cos 60.95 + j sin 60.95)
Express in polar form -14 - 7j
Tan θ = opp
___
Adj
Tan θ = -7
____
-14
θ tan-1 = - 7
______
-14
Θ = 26.56
√245 (cos 26.56 + j sin 26.56)
j^5 + j^ 3 - j^7
ReplyDeleteto approach this question youu will need to brake down j's to what you can easily undestand
eg.j^7= (j^2)x(j^2)x(j^2)x(j)
after you have done this for all you can work it out as a simple question
(√(-25) + j^3) + (j^7 - √(-9))
ReplyDeleteto approach this question, after you have brake into j^2's and j's only you need to know that:
(√(-25)= √(-1) x √(25)
after re write question and solve
4 + 7j added to 11 - 21j
ReplyDeleteto solve this question you simpily add the like terms
4 + 7j
+
11 - 21j
_________
15 - 14j
_________
.
j^5 * j^3
ReplyDeleteto approach this question you will need to like i did befor and brake down j's to what you can easily undestand
so.j^5= (j^2)x(j^2)x(j)
when you solve both then you can multiply
Ques1.j^5 + j^ 3 - j^7
ReplyDelete=j^5 + j^3 - j^7
=j^5 +j^3=j^8
= j^8 - j^7
= j
i thought that was the way to answer the question but i saw how boy-b work out the question and i think that is the correct solution
ReplyDeleteQues2.j^5 * j^3
ReplyDelete=i saw many different solutions to this question but this is how i would answer this question firstly taking the j^5 and put it in the form(j^2)(j^2)(j) and as fresh prince said then work out the j^3 in the ane way and then multiply them
1. j^5 +j^3 -j^7
ReplyDelete=(j^2 x j^2 x j) + (j^2 x j) - (j^2 x j^2 x j^2)
=j - j + j
=j
2. j^5 x j^3
ReplyDelete=(j^2 x j^2 x j) x (j^2 x j)
=j x -j
=-j^2
=1
7. (5 - 9j) x (6 - 3j)
ReplyDelete=30 - 15j - 54j + 27j^2
=30 - 69j - 27
=3 - 69j