- Evaluate (2 + j)(-1 +j/2)
- Evaluate (1 + 4j)^2
- Evaluate j(3 + 2j)(6 -4j)
- Give the modulus and argument of the following complex numbers
a)1 + j b)1 + 2j c) − 2 + 3j d)√2 − √(-36) e) - 3 - 5j - Evaluate a) (−1 + j)/(1 + j) b) j / (1 + 3j) c) (3 − 4j)/(12 + 8j) d) (2 − 3j)/(52 + 13j)
Friday, March 6, 2009
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Question1
ReplyDelete(2+j)(-1+j/2)
We first bring downstairs upstairs
So j/2 = 1/2j
(2+j)(-1+1/2j)
Now (a+b)(a+b)
Is a2 +ab+ab+b2
Therefore
-2+j-j+1/2j2
-2+1/2j2
But j2 = -1
So Ans = -2-1/2 = -5/2
question3
ReplyDeletefirst you multiply (3+2j)by j
= 3j+2j^2
but j^2= -1
therefore
(-2+3j)(6-4j)
-12+8j+18j-12j^2
as j^2= -1
then
-12j^2= +12
therefore
-12+8j+18j+12
-12+12+8j+18j
Ans= 26j
Question 1
ReplyDelete(2 + j) (-1 + j/2)
Now there are many ways to work this question, Goldfinger did the first and I will do a second.
Instead of bringing everything upstairs we can leave it as it is and expand.
(a + b) (a + b) = a^2 + ab + ab + b^2
This Implies
(2 × -1) + (2 × j/2) + (j × -1) + (j × j/2)
The 2 in the second brackets will cancel each other.
j × j/2 = j^2/2
The equation then becomes
-2 + j + (-j) + j^2/ 2
Recall j^2 = -1
The js will cancel one another
Now:
(-2) + (- ½) = -2 – ½
Answer = - 2 ½
Question 2
ReplyDelete(1 + 4j) ^2
This is the same as (1 +4j) (1+4j)
Now Recall (a+b) (a+b) = a^2 + ab + ab + b^2
Therefore: 1^2 + 4j + 4j + (4j) ^2
Simplify:
1 + 8j + 16j^2
Recall j^2 = -1
This Implies, 1 + 8j + (-1) (16)
Therefore,
(1 – 16) + 8j = -15 + 8j
recall: √-1 = j and j^2 = -1
ReplyDeleteQuestion 3. j(3+2j)(6-4j)
get rid of the brackets by expanding;
taking j(3+2j) => 3j + 2j^2 => this can be broken down into:
3j + 2(-1) => -2 + 3j
then multiply (-2+3j) (6-4j) = -12 +8j +18j -12j^2
simplify: -12 +26j -12(-1)
answer = 26j
NB; when expanding brackets remember to use ur loops.
A 10 ohm resistor in series with a 2 microfarad capacitor is connected parallel to a 20 ohm resistor in series with a 0.5 H inductor. If the source voltage is 24volts and the frequency is 50hz, find the total impedance (Z) and the phase angle.
ReplyDeleteQuestion 3
ReplyDeletej (3 + 2 ) ( 6 – 4j )
First we work out
j ( 3 + 2)
We then get,
(3 × j) + (2 × j) = 3j + 2j
Now the equation becomes
(3j + 2j) (6 – 4j)
Recall (a + b) (a + b) = a^2 + ab + ab + b^2
So the result is then
18j – 12j^2 + 12j – 8j^2
= 18j + 12j – 12(-1) -8(-1)
= 18j + 12j + 12 + 8
= 30 j + 20 ( final answer)
Question 3 Alternate Method
ReplyDeletej ( 3 +2 ) ( 6 – 4j )
We can expand the brackets (ignoring the j)
(3 + 2) (6 – 4j)
= 18 + 12 – 12j – 8j
Now we multiply this new equation by j
j ( 18 + 12 – 12j – 8j )
= 18j +12j – 12j^2 – 8j^2
= 18j +12j – 12 (-1) – 8 (-1)
= 18j + 12j + 12 + 8
Answer = 30j + 20
QUES 4
ReplyDeletethe mudulus is when u sqaure both terms(imag and real) add them and find the sqaure root for it
the argument is u find the angle, well am not quite sure how to explain it, but using this fomula sure help
tan=imag/real
and u make the subject of the formula to find the angle
A)√(1)^2+(1)^2
=√2
arg tanQ=1/1
Q=45degrees +ve 1st quadrent
B)√(1)^2+(2)^2
=√5
arg tanQ=2/1
Q=63.4 degrees +ve 1st quadrent
C)√(-2)^2+(5)^2
=√29
arg tanQ=5/2
=68.2 because in -ve 2nd quadrent
Q=180-68.2
Q=111.8 degrees
D)√2-√(-36) = √(2)^2+(-6)^2
=√40
arg tanQ=6/2
=71.6 because in -ve 4th quadrent
=360-71.2
Q=289 degrees
E)√(-3)^2+(-5)^2
=√34
arg tanQ=5/3
=59 because in +ve 3rd quadrent
Q=180+59
Q=239 degrees
ghetto celeb did u put that ques up to get some help with it ???
ReplyDelete2)
ReplyDelete(1 + 4j)^2
we bring it to simpler terms by removing the square, so we don't get confused when multiplying;
(1 + 4j) (1 + 4j)
1 + 4j + 4j +16j^2
1 + 8j + 16(-1)
8j - 15
3) j(3 + 2j)(6 -4j)
ReplyDeletefirst thing, we expand the brackets,
(3j + 2j^2) (6 -4j)
after expanding the first term, i placed it back in brackets, so ill know what im multiplying;
18j + (-12j^2) + 12j^2 - 8j^3
18j - 8j^3
10j^2
10(-1)
-10
yes i did put the question up to get some help
ReplyDeleteQUES 5
ReplyDeletefirst things first when wroking these types of questions, you have to look out for two things. one is the police. make sure when multipling or adding you use your calculators and second thind is the upstairs and down stairs. when u see questions like these you have to find what is called the condjrigate. This is that the value that is in the denominator and sitch the sign that is in front of the imaginary trem. then you multiple out the trems or expand the brackets. you will end up having no imanginary trems in the denominator.
A) (−1 + j)/(1 + j) X (1 - j)/(1 - j)
= -1+j+j+1 / 1-j^2 j^2=-1
= 2j/2 = j
B) j / (1 + 3j) X (1 - 3j)/(1 - 3j)
=j+3 / 1-9j^2 j^2=-1
=j+3 /1+9
=j+3/10
C) (3 − 4j)/(12 + 8j) X (12 - 8j)/(12 - 8j)
=36-24j-48j+32j^2 / 144-64j^2 j^2=-1
= 36-24j-48j-32 / 144+64
= 4 - 72j / 208
D) (2 − 3j)/(52 + 13j) X (52 - 13j)/(52 - 13j)
=104-26j-156j+39j^2 / 2704-169j^2
=65-182j / 2873
How to answer this question
ReplyDelete'(5 + 3j)^2
2 answer yur question darky..
ReplyDeletealways try to remove ur powers if u could.
this makes it easer 4 d eys to see.
recall j^2 = -1
(5+3j)(5+3j)
= 25 + 15j + 15j + 9j^2
= 25 + 25j + 9(-1)
= 25 + 25j -9
= 25 - 9 + 25j
= 16 + 25j
wel darky
ReplyDelete(5 + 3j)^2
is the same as saying
(5 + 3j)*(5 + 3j)
so this means you multiply everything in equation 2 by 5
(5*5)+(5*3j)
Then you multiply everything in equation 2 by 3j
(3j*5)+(3j*3j)
now add the to parts together and you get
(5*5)+(5*3j)+(3j*5)+(3j*3j)
and that is equal to
=25+15j+15j+9j^2
and remember that j^2=-1
so this means
=25+30j+(9(-1))
=25+30j-9
=16+30j
and zipper 15j + 15j is equal to 30j not 25j
ReplyDeleterenjisan if my eyes aren't playing a trick on me, for ques 4.
ReplyDeletePart C asked for
-2+3j not -2+5j
which wud be
(-2^2)+(3^2) all to be squarerooted
which is equal to
=√(4+9)
=√13
and in Part D
you were inconsistant wen you square rooted your terms. you said that
√2-√(-36) = √(2)^2+(-6)^2
=√40
i dont tink that so
cuz you also suppose to square root 2
so i think the correct working is suppose to be
√2-√(-36) = (√2^2)+(√(-36)^2)
and that is equal to
=2+36
=38
wel i think thats the correct ansa,
any1 wants to confirm my ansa?
no 2
ReplyDelete(1 + 4j)(1 + 4j) = 1^2 + 4j + 4j + 16^j^2
= 1 + 8j -16
= -15 + 8j
no 2
ReplyDeletej(3+2j)(6-4j)
to work this out we have to work the bracket first:
j(18-12j+12j-8j^2) = j(18+8)
= 26j
hey cold some one please tell me what is meant by the word argument and what are they asking you to find?
ReplyDeleteEvaluate (1 + 4j)^2
ReplyDeletethat is the same as
(1+4j)(1+4j)
so this = 1+4j+4j+4j^2
an j^2 =-1
-3+8j
ridiclyric you mixed up your answer for question 2. The correct way to do it is how it is done above in Blade's answer. Please review it and practice another one like it
ReplyDeleteThe modulus is basically squaring the real and the imaginary and adding dem and the the root ..for dese kind of questions u hav to look out for police and downstairs...
ReplyDeleteI suppose the word arguement in the question was for you to explain different ways of answering the question.
ReplyDeleteWho's arguing "low rider".....?
ReplyDeleteQuestion3
ReplyDeleteEvaluate j(3 + 2j)(6 -4j)
firstly you open out the brackets (3+2j)by j
= 3j+2j^2
but j^2= -1
therefore
(-2+3j)(6-4j)
Now we open out the brackets again
-12 + 18j + 8j +12j^2
but j^2= -1
then
-12 + 18j + 8j + 12(-1)
therefore
-12+8j+18j-12
-24+26j