Calculate
(a) (2 − 14j) + (15 + 4j) (b) (12 − 8j) − (5 + 14j) (c) (2 − 6j) × (3 + 2j)
(d) (3 − 4j) × (7 + 2j) (e) (5 − √(-16)) × (2 + √(-9)) (f) (1 +2 j)^2
Plot 1 + 6j, 2 − 9j, −9 +7 j and −11 − 21j on the Argand diagram and for each write as polar coordinates
Calculate the modulus and phase angle of 1 + 12j, 2 − 12j, −14 + j and −5 − 9j.
Solve the quadratic equations
(a) x^2 − x + 5 = 0;
(b) 2x^2 + x + 8 = 0;
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What is meant by the Impedance ?
ReplyDeletei think impedance is the total resistance of a circuit that has resistors ,capacitors and inductors. it is usually associated with a.c voltages
ReplyDeletequestion b:
ReplyDelete(12-8j)-(5+14j)
First of all you must remember that j= square root -1, and j^2= -1
You then look for any term within brackets that is raised by a power
you then put the real terms under the real terms and the imaginary under imaginary
12 - 8j
- 5 + 14j
----------
= 7 - 22j
that would be how to work that problem out
yea but if is in a parallel cirt you will hav to find out wot is Z1 and Z2.The imaginary well be Xl -Xc and den after yuh subtitute in the formula Z1Z2 upone Z1+Z2 and den my frend youll find ZT!!!!!!!
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteyeah people how to cnvert to polar form????
ReplyDeletehow u mean how to concert d polar form... u jus open ur brackets an multiply it by "r"... i tink lol
ReplyDeletea/ 2-14j
ReplyDelete+
15+4j
-----
17-10j
------
b/ 12-8j
_
5=14j
-----
7-22j
-----
c/ (2-6j) (3+2j)
6+4j-18j+12j^2
=18-12j
d/(3-4j) (7+2j)
21+6j-28j-8j^2
21+8+6j-28j
=29-22j
f/ (1+2j)^2
(1+2j) (1+2j)
1+2j+2j+4j^2
=(-3)+4j
sombody answer part e/ in d frist question NA!!!!!!!!!!!
ReplyDelete(5 − √(-16)) × (2 + √(-9))
ReplyDelete(5- √16 j) (2+ √9 j)
(5-4j) ( 2+3j)
and u den multiply normal wit d conj and ect
i kno some1 asked about the polar form b4 but can any1 explain it a little more detailed?
ReplyDeleteyes SMILEY and crazykid
ReplyDeletethere is no convertin tuh polar form, well not really say for instants u have 3+4j right!! real term and imag temr right!! What the poler form does is to show that same expression w.r.t cos and sin
where cos is the real term
and
sin is the imag term
most time they only ask u this when u have tuh find the argument
OR
once they ask u to show in the polar form
E.G
3+4j
becomes
3cos + 4sinj
do u all understand jus let me no and i'll try to explain it more since some ppl might not understand what i wirte
for u to understand the ques smiley jus remember that
ReplyDeletej^2=-1
that is very important when doing complex no. and barckets i'll work the ques for u
E) (5 − √(-16)) × (2 + √(-9))
(5 − √-1 x √16) × (2 + √-1 x √9)
(5 − √j^2 x √16) × (2 + √j^2 x √9)
(5 − j x √16) × (2 + j x √9)
(5 − j x 4) × (2 + j x 3)
(5 − 4j) × (2 + 3j)
then from here u can expand the brackets
(5 − 4j) × (2 + 3j)
10 + 15j - 8j - 12j^2
10 + 7j - 12 x -1
10 + 7j + 12
22 + 7j
thats your final answer and ill show you what is that polar form as well
in polar form it will be
22cosQ + 7sinQj
and i'll leave the angle for u to find
Q represents the angle right?
ReplyDeleteso how exactly to find that now?
ReplyDeletesomeone pls ansa (:(
quadratic for x^2-x+5=0
ReplyDeletehuh i really cant remember, my brains getting rusty..lol.
well for x^2-x+5=0
ReplyDeletefirst you remember the quadratic equation
x = -b (+or-) √b^2 - 4ac all divided by 2a
note that in the quadratic a = x^2 coefficient, b = x coefficient and c = x^0(which is 1) coefficient
after you substitute the values of a, b, and c into the equation and you can work it from there
according to renjisan to find Q(i.e. the angle) the following is as follows (lol)
ReplyDeleteremember that miss taught us that tanθ=opp divided by the adj, well when the graph is drawn to represent the complex function, and the angle is required to be found by the acquirement of a triangle, the angle can be found by
tanθ=imaginary divided by the real
BUT the angle we want is not that of the angle found by the equation
tanθ= imaginary divided by the real
but the total amount of degrees in a circle(360) minus the angle found
oh and i forgot to say hi to all, i missed the blog cuz i got dismembered somehow lol
ReplyDeleteany ways could anyone help me with a question like this
(5 − √(-16)) + (2 + √(-9))
then divided by 2 + √(-25)
to me all complex nos is the same thing and that all there is to know is that j^2=-1, j=-1^1/2
ReplyDeleteand never to work out complex nos IN YOUR HEAD (not recommended), use a god damn calculator check this -6*-25=?
and
i got that wrong sucks doesn't it especially if it 5 marks
regarding fariel'sbeststudent
ReplyDeletefirst you remember that j=√-1 and remember de police
5 - √(-1x16) + 2 + √(-1x9) divided 2 + √(-1x25)
5 - √(-1)√(16) + 2 + √(-1)√(9) divided 2 + √(-1)√(25)
i am stuck can some one help
me please
you need to take it step by step
ReplyDeletesomeone else elaborate and help both damon and ME
(5 − √(-16)) × (2 + √(-9))
ReplyDeletethe first step in solving this problem is to solve in the brackets first
(5-√16j)x(2+√9j)
this is because j=-1
the second step is to find the √ of each term
(5- 4j)x(2+3j)
the next stem is to remove the brackets by expanding
(5 − 4j) × (2 + 3j)
10 + 15j - 8j - 12j^2
10 + 7j - 12 x -1
10 + 7j + 12
22 + 7j
so for any complex number question given we can solve it by using the values for j^2 and j, rite?
ReplyDeleterecall: complex numbers consist of rel and imaginary terms. the imaginary terms are denoted by the letter j.
ReplyDeletein the addition of complex numbers, the real are added to together and the imaginary are added toether:
(a) 2 - 14j +
15 + 4j
=17 - 10j
the subtractio of complex numbers are similar to th addition in that the real are subtracted from the real and the imaginary are subtracted from the imaginary:
(b) 12 - 8j -
5 + 14j
= 7 - 22j
when multiplying complex numbers the first term in the first bracket is multiplied by the terms in the second bracket, then the second term of the first bracket is multiplied by the terms in the second brackets:
(c) (2-6j) (3+2j)
= {(2*3)+(2*2j)} - {(6j*3)+(6j*2j)}
= {6+4j} - {18j+12j^2}
=6+4j-18j+12j^2
recall:j^2=-1
= 6+4j-18j+12(-1)
all real terms are added and all imaginary terms are added:
= 6-12+4j-18j
=-6-14j__________check back ur ans 4 dis smiley
(e) 5-√-16
recall: j=√-1
therefore √-16= √-1 * 16=
= j√16
= 4j (√16 =4)
5-√-16 = 5-4j
hope u understand (e) smiley, ow can any1 please tell how to plot on argand cart
ReplyDeleteimpedence(Z)is the total resistance to the flow of current by the resistors, capacitors and inductors in the circuit.
ReplyDeletehe value of the resitor is the real and the vlue of the difference of the inductor and the capacitor is te imagiary
sorry i dident finish the question, (is weekend na)
ReplyDeletee) 5-√-16
recall: j=√-1
therefore √-16= √-1 * 16=
= j√16
= 4j (√16 =4)
5-√-16 = 5-4j
now we deal with 2+ √-9
√-9= √-1*9 = j√9 = 3j
2+ √-9 = 2+3j
now we multiply: (5-4j) (2+3j)
= 5(2+3j) - 4j(2+3j)
=10+15j-8j-12j^2 (rem j^2= -1 so 12j^2= -12)
= 22+7j
OK fariel'sbeststudent
ReplyDeletethe question you asked
(5 − √(-16)) + (2 + √(-9))
then divided by 2 + √(-25)
well the first thing u do is work out the top that is (5 − √(-16)) + (2 + √(-9)) first and from above the answer that u will get is 22 + 7j
then u work out the bottom
2 + √(-25)
2 + √-1 x √25
2 + √j^2 x √25
2 + j x 5
2 + 5j
then u divide it by placing the top part u worked out over the bottom part u worked out
so it will be (22 + 7j)/(2 + 5j)
(22 + 7j)/(2 + 5j) X (2 - 5j)/(2 - 5j)
when expanded you get
(44 - 110j + 14j - 35j^2)/(4 - 25j^2)
(44 - 110j + 14j - 35 x -1)/(4 - 25 x -1)
(44 - 110j + 14j + 35)/(4 + 25)
79 - 96j/29
and that is what u will get for your answer
well animechic it depends on how the question is stated
ReplyDeletefor example if they give u something like this
(5 − √(-16)) + (2 + √(-9))
then u can use j^2=-1 to get it into the complex form and if the ques reqires you to find the modulus or argument
and the other way is they can give u the impedence of resistoer, capacitors or inductors and tell to find the phase angle or total impedence
i think ken should hav gone a step further.
ReplyDelete(5 − √(-16)) × (2 + √(-9))
the first step in solving this problem is to work out the square root part in the brackets first.
(5-√-1*√16) x (2+√-1√9)
since j = √-1
(5-√16j) x (2+√9j)
the second step is to find the √ of each term
(5- 4j)x(2+3j)
the next step is to remove the brackets by expanding
(5 − 4j) × (2 + 3j)
10 + 15j - 8j - 12j^2
Since J2 = -1
Then 10 + 15j -8j –12(-1)
10 + 7j - 12 x -1
10 + 7j + 12
=22 + 7j
Calculating modulus and phase angle...
ReplyDeletereal + imag
1 + 12j...
The first thing to remember is to plot the points. The real numbers goes on the x-axis and the imaginary nos goes on the y-axis. The graph plotted will form a right angle triangle.
The modulus will be the hyp of the triangle.
recall => hyp = √opp^2 + adj^2
therefore taking 1 + 12j
modulus = √(1^2) + (12^2)
= √1 + 144
The phase angle is the angle between the x-axis and the hyp.
The phase angle is calculated using ;
(let a = the phase angle)
tan a = opp/adj
tan a = 12/1
a = tan^-1 (12/1) = 85.3 degrees => phase angle...
Polar coordinates...
ReplyDeleteTo put complex nos into polar the first thing you need to do is plot the point/s given. Therefore using 1 + 6j, where 1 is the real no. (goes on the x-axis) and 6j is the imaginary (goes on the y-axis)
Then find the modulus => √opp^2 + adj^2
= √(1^2) + (6^2) = √1 + 36
= 6.08
and the phase angle = tan a = opp/adj
a = tan^-1 (6/1) = 80.5 degrees
Put in the polar form => r(cos a + j sin a)
where r is the modulus and a is the phase angle.
Ans => 6.08(cos 80.5 + j sin 80.5)
heyy i just wanted to point out something important...when calculating the phase angle...always remember to check which quadrant of the graph it falls in because in the 2nd quadrant or the bottom half of the right side of the graph the phase angle is calculated by...
ReplyDelete360 - [tan^-1 (opp/adj)]
..this is something we might forget in the rush so i just thought i'd remind u :)
1 + 12j
ReplyDeletethe phase angle is:
tan x = opp/adj
tan x = 12/1
= 85 degrees
the modulus is:
r = (12^2 + 1^2)^1/2
= 12
2- 12j
ReplyDeletethe modulus is:
r = (2^2 + (-12^2))^1/2
= 12
the phase angle is:
tan x = opp/adj
tan x = 12/2
x = 81 degrees
since the phase angle is in the 4th quadrant it will be:
360 - 81 = 279 degrees
-14 + j
ReplyDeletethe modulus is:
r = ( -14^2 + 1^2)^1/2
= 14
the phase angle is:
tan x = opp/adj
x = 1/14
x = 4
since the phase angle is in the 2nd quadrant it will be:
180 - 4 = 176 degrees
-5-9j
ReplyDeletethe modulus is:
r = ( -5^2 + -9^2)^1/2
= 10
the phase angle is:
tan x = opp/adj
tan x = 9/5
x = 61
since the phase angle is in the 3rd quadrant it will be:
180 + 61 = 241 degrees
x =
when you have a question like:
ReplyDelete(2-14j)+(15+4j)
2 and 15 are considered to be real numbers
and -14j and 4j are considered to be imaginary numbers.In order to simplify the equation u must group the like terms real with real and imaginary with imaginary.
2-14j
+15+ 4j
=17-10j
(2 − 14j) + (15 + 4j)
ReplyDelete= 17 - 10j
r = 10^2 + 17^2 = root of 389
tan teta = 10/17
= 30.5
17- 10 j = root 389 ( cos 30.5 + j sin 30.5)
a) (2 − 14j) + (15 + 4j)
ReplyDelete= 17 - 10j
(b) (12 − 8j) − (5 + 14j)
= 8 - 22j
Miss tried to flow us with those minus signs, lol. Come on miss, try harder
c) (2-6j) (3+2j) we open out the brackets
ReplyDelete=6+4j-18j+12j squared (we know that j squared is equal to minus 1 right people)
therefore.
=18-12j
d)(3-4j) (7+2j) we open out the brackets which gives
= 21+6j-28j-8j squared
= 21+8+6j-28j
=29-22j
f) (1+2j)squared is simply
=(1+2j) (1+2j)
=-1+2j+2j+4j squared
=(-3)+4j
And for the person above who asked about polar form, remember after plotting the graph of the real against imarginery and finding the angel of impedencem, the polar form will be
sqaure root r [cos(the angle) + j sin(the angle]. Correct me if am wrong
That is correct what you are saying because the sqaure root r is just the squares of real and imajenry the angle may be found using tan teta .
ReplyDeleteHey swanky thing. your asnwer for part c is wrong
ReplyDeletec) (2-6j) (3+2j) --- using the looping method
= 6+ 4j-18j-12j2 ------ since j^2 = -1
=6 + 4j – 18j – 12(-1)
=6 – 14j +12
=18 – 14j
(5 − √(-16)) × (2 + √(-9))
ReplyDelete[(5*2)+(5*√(-9)] - (√(-16 *2) -(√(-16*√(-9)
making √(-9) = √(9)*√(-1) where √(-1) = j
therefore 3j
and √(-16) = √(16)*√(-1) where √(-1) = j
therefore 4j
substitute into
[(5*2)+(5*√(-9)] - (√(-16 *2) -(√(-16*√(-9)
where
[(5*2)+(5*3j)] - (4j *2) -(4j*3j)
=10 + 15j - 8j -12j^2
=10 + 7j + 12
=27 + 7j
Thanks for the correction Small Man, i must have mixed up myself in the process
ReplyDeleteIn the question (3-4j)(7+2j)
ReplyDeletetaking into consideration when the brackets are expanded and multiplied the like terms are grouped together using loops
3(7+2j)-4j(7+2j)
=21+6j-28j-8j
=21-22j-8j
=21-30j
(d) (3 − 4j) × (7 + 2j)
ReplyDeletefirstly expand the brackets
=21+6j-28j-8j^2......sincej^2=-1
=21-22j+8
=29-22j
can someone show me how to calculate the phase angle and the modulus please.......
ReplyDeleteokay phase angle is calculated by saying tan teta = opp/ adj
ReplyDeleteQuestion 1
ReplyDelete1 + 12j
the phase angle you use tan teta = opp / adj
tan teta = opp/adj
tan teta = 12/1
= 85 degrees
the modulus is:
r = (12^2 + 1^2)^1/2
= 12
Set of 12
ReplyDelete(a) (2 − 14j) + (15 + 4j)
Work imaginary numbers together and real numbers together.
2 + 15 = 17
-14j + 4j = 10j
(b) (12 − 8j) − (5 + 14j)
12 - 5 = 7
-8j - (+14j) = -22j
(c) (2 − 6j) × (3 + 2j)
(2 x 3) - (6j x 3) + (2 x 2j) - (-6j x 2j)
6 - 18j + 4j + 12j2
6 - 14j - 12
-6 – 14j
(d) (3 − 4j) × (7 + 2j)
(3 x 7) - (4j x 7) + (3 x 2j) + (-4j x 2j)
21 - 28j + 12j - 8j2
21 - 16j + 8
29 – 16j
(e) (5 − √(-16)) × (2 + √(-9))
Since we cannot find the square of negative numbers the approach is.
( 5 - √-1 x 16 ) (2 + √-1 x 9)
(5 - 16 j) x (2 + 9j)
(5 x 2) - (16j x 2) + (5 x 9j) + (-16j x 9j)
10 - 32j + 45j - 144j2
154 - 13j + 144
154 – 13j
(f) (1 +2 j)^2
(1 +2 j) (1 +2 j)
(1 x 1) + (2j x 1) + (1 x 2j) + (2j x 2j)
1 + 2j + 2j + 4j2
1 + 4j - 4
-3 + 4j
(6 - 7j) (6 - 7j) x (4+ 3j)
________ _______ _______
(4 - 3j) 4 - 3j) (4+ 3j)
4^2 - 3j^2
16 + 3^2
25
1/25 (6 - 7j) (4 + 3j)
1/25 (6 x 4) - (7j x 4) + (6 x 3j)+(-7j x 3j)
24 - 28j + 18j - 21j2
24 - 10j + 21
45 – 10j
(d) (3 − 4j) × (7 + 2j)
ReplyDeletethe first step is to multiply each term to remove the brackets
(3 x 7) - (4j x 7) + (3 x 2j) + (-4j x 2j)
21 - 28j + 12j - 8j2 multiply 8x -1
21 - 16j + 8
29 – 16j
1 + 6j
ReplyDeletethe polar cordinates s in the form
r(cosx + j sin x)
in complex nos. what u have to remember is that there are imaj, and real nos.
ReplyDeletethe real does not have the j
(2-14j) + (15+4j)
ReplyDeleteyou identify the real nos, 2 and 15
then the imajinery has the j so they are -14 and 4.
NOTE: REMEMBER TO LOOK OUT FOR POLICE
remember for polar coordinates it would be
ReplyDeleter(cos+jsin theta)