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Thursday, March 5, 2009
complex numbers questions 3
What is the magnitude of √(-8)? What is the magnitude of √(-14)? What is the magnitude of 5 + 6j? Is this real or imaginery √(-16)? What is the phase angle of 7 + 9j? What is the magnitude of 5 + 6j?
when doing these questions always remember that the square root of a negative number is an imaginary number, and can be broken up into parts jus as joh said eg. √(-14) can be written as √-1*14 which can be further simplified by √-1 √14. as you know the √-1=j therefore you will finally obtain √14j.
to calculate the phase angle (@) you must use the formula (tan@=imaginary/real, therefore for the equation 7+9j, 7 is the real number and 9j is the imaginary number. using the formula and substituting you will get tan@=9/7 therefore @=tan^1 9/7= 52.13 degrees.
to calculate the magnitude of 5+6j you have to use pythagoras' theorem which is a^2= b^2 + c^2. to make (a) the subject of the formula you will get a= √b^2 + c^2, then you substitute the equation into the formula and get lAl= √5^2 + 6^2 = 7.81
wow angel dat was a very gud explaination.... Wel i hav a question....
For a circuit in which the resistor is 8.00ohm,the inductor is 7.00ohm, and the capacitor is 13.0ohm, find the impedence and the phase angle between the current and the voltage.
My mistake, i didn't visualize the graph so in the case of the phase angle, thank you computation for highlighting my error......it will be the opp over adj which is 9 over 7.....
Phase Angle What exactly is the phase angle? The phase angle is known to be the Tan-1 of the imaginary part divided by the real i.e. Tan -1 (imag / real)
Therefore the Phase Angle of 7 + 9j is simply
Tan Ө = 9j (imag)/ 7 (real)
Ө= Tan-1 (9j/7)
At this point it is quite easy to substitute for j and solve.
(Q2) {√ (-25) + √ (64)} ∕ {√ (81) + √ (-9)} let’s call √ (-25) + √ (64) _ X and √ (81) + √ (-9) _ Y to divide X by Y , we have to multiply the numerator, X and the denominator, Y by the conjugate of Y. Two complex numbers of the form a+bj and a-bj are said to be conjugate
To find the magnitude of any given values u use Pythagoras's theorem...and apply by knowing dat the real goes on the x axis and the imaginary on the y axis
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√(-8) = √(-1*8) = √-1*√8
ReplyDeletesince √-1 = j
therefore answer = j√8
the magnitude of root(-14) =root-1*root14
ReplyDeletewhich equals root14j
where j is root -1
jade how do you get the square root sign
ReplyDeletethe squareroot of -16 is an imaginary number. i would also like to know how do u get the squareroot sign!!!!
ReplyDeletewhen doing these questions always remember that the square root of a negative number is an imaginary number, and can be broken up into parts jus as joh said eg. √(-14) can be written as √-1*14 which can be further simplified by √-1 √14. as you know the √-1=j therefore you will finally obtain √14j.
ReplyDeleteyou jus follow the same format and you will be able to find the magnitude of √(-8).
ReplyDelete√(-16)is imaginary, because as i said befor from what i know the square root of a negative number is imaginary.
ReplyDeleteto calculate the phase angle (@) you must use the formula (tan@=imaginary/real, therefore for the equation 7+9j, 7 is the real number and 9j is the imaginary number. using the formula and substituting you will get tan@=9/7 therefore @=tan^1 9/7= 52.13 degrees.
ReplyDeleteto calculate the magnitude of 5+6j you have to use pythagoras' theorem which is a^2= b^2 + c^2. to make (a) the subject of the formula you will get a= √b^2 + c^2, then you substitute the equation into the formula and get lAl= √5^2 + 6^2 = 7.81
ReplyDeletewow angel dat was a very gud explaination....
ReplyDeleteWel i hav a question....
For a circuit in which the resistor is 8.00ohm,the inductor is 7.00ohm, and the capacitor is 13.0ohm, find the impedence and the phase angle between the current and the voltage.
This comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteIn the case of √ (-16) it is an imaginary number, because it is the square root of a negative number.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteWhat is the phase angle of 7 + 9j?
ReplyDeletewell the phase angle is easy to find it just as how miss showed us in class ........
tanӨ= opp./adj.
=ima./real
tanӨ = 9/7
Ө=tan^1 9/7
Ө=52.12 degrees
well calculating the magnitude of 5+6j is quite easy following the proceedures that we have learnt in class to work it out....
ReplyDeletemagnitude of 5+6j
using pythagora's theorem...
sub. the eqn. into the formular
you will get......
=√5^2+6^2
=7.81
My mistake, i didn't visualize the graph so in the case of the phase angle, thank you computation for highlighting my error......it will be the opp over adj which is 9 over 7.....
ReplyDeleteimaginery over real people, imaginery over real....phase angle
ReplyDeletePhase Angle
ReplyDeleteWhat exactly is the phase angle? The phase angle is known to be the Tan-1 of the imaginary part divided by the real i.e. Tan -1 (imag / real)
Therefore the Phase Angle of 7 + 9j is simply
Tan Ө = 9j (imag)/
7 (real)
Ө= Tan-1 (9j/7)
At this point it is quite easy to substitute for j and solve.
Same method would apply for 5+ 6j.
What is the magnitude of 5 + 6j?
ReplyDelete-First square both nos.
5^2+6^2
-secondly find the square root of these squared nos.
√5^2+6^2
√25+36
Ans. =√61
i think that the √(-16)is imaginary because any number that contains j is imaginery and √(-16)works out to be :
ReplyDelete= √-1x16 = √-1 √16 = √16 j = 4j which is imaginary
the magnitude of (-8)^1/2 = (-1 x 8)^1/2
ReplyDelete= (8^1/2) j
the magnitude of (-14)^1/2 is done the same way as the problem above:
ReplyDelete= (-1x14)^1/2
=(14^1/2) j
the magnitude of (5 + 6j) is calculaated using pythagoras theorem:
ReplyDeleter = (5^2 + 6^2)^1/2
= 7.8
(-16)^1/2 = (-1 x 16)^1/2
ReplyDelete= (16^1/2) j
because this expression contains j it is imaginary!!!
we will use trigs to calculate the phase angle for (7+9j):
ReplyDeletetan x = opp/adj
tan x = 9/7
phase angle = 52.1 degrees
the last question is the same as the third. however even if the values were different, magnitude can always be found using pythagoras theorem:
ReplyDelete(Q2) {√ (-25) + √ (64)} ∕ {√ (81) + √ (-9)}
ReplyDeletelet’s call √ (-25) + √ (64) _ X
and √ (81) + √ (-9) _ Y
to divide X by Y , we have to multiply the numerator, X and the denominator, Y by the conjugate of Y.
Two complex numbers of the form a+bj and a-bj are said to be conjugate
Numerator: √ (-25) + √ (64) → (√ -1 * 25) + 8
= j √ 25 + 8
= 5j + 8
Denominator: √ (81) + √ (-9) → 9 + (√ -1 * 9)
= 9 + j √ 9
= 9+ 3j
X/ Y : 5j + 8/9+ 3j____________ the conjugate of Y is 9- 3j
Numerator (5j + 8) (9- 3j) = 87+ 2j
Denominator (9+ 3j) (9- 3j) = 90
Ans. 1/90 (87 +21j)
this is for anothe question sorry
ReplyDeletefor Q1 i thought the magnitude wod of been jus 8?
ReplyDeletethis phase angle thing is new to me.
ReplyDeleteis there imaginery numbers in real life?
mysticwings how did you get 52.1 degrees for
ReplyDeletetanx=9/7?
ok i got the answer but why is the unknown angle 1 for tanx?
ReplyDeleteboy-b i dont really understand what you are asking, can you plz explain??
ReplyDeleteWhat is the magnitude of 5 + 6j?
ReplyDeleteto find the magnitude use Pythagoras theorem
x^2= 5^2+6^2
x= square root of
x= 7.81
To find the magnitude of any given values u use Pythagoras's theorem...and apply by knowing dat the real goes on the x axis and the imaginary on the y axis
ReplyDeleteWhat is the magnitude of 5 + 6j?
ReplyDeletemagnitude = root(5^2 + 6^2)
= root 61
As a reminder when checking the phase you check from the anti-clockwise direction after finding tan.
ReplyDeleteWhat is the magnitude of √(-8)?
ReplyDeletefirst you must remember that the root of -1 is j
then you say √(-1*8) = √-1*√8 =
j√8
What is the magnitude of √(-14)?
ReplyDeleteI think it the approach is -1
*14 ^1/2
=14^1/2 j
remeber to put in your j
What is the magnitude of 5 + 6j?
ReplyDelete5^2 + 6^2 ^1/2
= 7.8
# 4
ReplyDelete√(-16)
firstly, u simplify
√16√-1
find the sq. rt. using=> √-1 = j
=> 4j
therefore it is definitely imaginary
#7
ReplyDelete(5 - 3j)(6 - 4j)
firstly, expand the brackets
so u'll get=> (5*6)+(-3j*6)+(5*-4j)+(-3j*-4j)
30 + (-18j)+ (-20j)+ 12j^2
use j^2 = -1: 30 - 38j+ 12(-1)
30 -38j - 12
18 - 38j
How do u find the magnitude??
ReplyDeleteWhat is the magnitude of √(-8)?
ReplyDeletejust concider that
√(-8) = √(-1)x √(8)
there for
√(-8) = 2.8j
What is the magnitude of √(-14)?
ReplyDeleteas i did before concider that
√(-14) = √(-1)x √(14)
there for
√(-14) = 3.7j