- Evaluate (2 + 3j) - (6 –4j)
- Evaluate (2 + 3j)(1 – 5j)
- Write the conjugate of a) 3 + 5j b) 2 – 6j
- Evaluate (3+ 4j)/(3-2j)
- Evaluate [3 + √(-25) ] + [ 8 - √(-16)]
- Evaluate [3 + √(-25) ] - [ 8 - √(-16)]
- Evaluate [3 + √(-25) ] x [ 8 - √(-16)]
- Evaluate [3 + √(-25) ] / [ 8 - √(-16)]
Friday, March 6, 2009
Subscribe to:
Post Comments (Atom)
(2+3j)-(6-4j)
ReplyDeleteas i always say real under real imaginary under imaginary, and look out for yuh know who, yes the police who is mister minus sign.
you will get for the real 2-6=-4
and for the imaginary 3j-(-4j). rememba a minus and a minus gives a plus so you will get
3j+4j=7j therefore the answer will be -4+7j.
plz anyone if im wrong plz correct me. so il know my mistakes, tanks.
ReplyDeleteyea it looks correct to me...
ReplyDeleteEvaluate (2 + 3j)(1 – 5j)
ReplyDeletehmm...wel multiply the first term in the first bracket with al those in the second bracket,so it wil b
(2X1)+(2X-5j)=2-10j and the second term in the first bracket wit all the terms in the second,
(3jX1)+(3jX-5j)= 3j-15j^2
NOW place all terms 2gether:
2-10j+3j-15j^2
2-7j-15j^2
2-7j-15(-1) (since j^2 is -1)
2-7j+15
2+15-7j
17-7j
if im not mistaking the conjugate is the opposite of what is given, for example 9-6j, the conjugate of that is 9+6j, as you can see the minus sign changed to a plus which is the opposite. therefore the conjugate for question 3a) 3 + 5j b) 2 – 6j are as follows:
ReplyDelete3a) the conjugate of 3+5j is 3-5j
b) the conjugate of 2-6j is 2+6j
as you can see the sign change and is just the opposite.
question 4.
ReplyDelete(3+4j)/(3-2j)
when deviding you multiply by the conjugate of the denominator. the denominator is 3-2j therefore its conjugate is 3+2j.
so you'll get 3+4j/3-2j X 3+2j/3+2j
working numerator first.
(3+4j)(3+2j)
you simply expand like a normal polynomial and you'll get (3x3)+(3x2j)+(4jx3)+(4jx2j)=9+6j+12j+8j^2
=9+18j+8j^2
but remember j^2=-1
so you have to substitute this into the equation and you'll get 9+18j+8(-1)=9+18j-8
=9-8+18j
=1+18j
working the denominator
(3-2j)(3-2j)
using the concept of (a+b)(a-b)=a^2-b^2
where from the equation a is 3 and b is 2j
so you'll get 3^2-2j^2=9-2j^2
but remember again j^2=-1
so you'll get 9-2(-1)=9+2=11 (remember a minus by a minus gives a plus)
therefore you'll get 1/11 [1+18j] or 1+18j/11, which in real life deals with electricity and it represents the impedence in the circuit.
Evaluate [3 + √(-25) ] / [ 8 - √(-16)]
ReplyDeleteFirstly √(-25)can be broken up into √-1 an √25 which can futher break down to get 5j since
√-1 is j^2 an √25 is 5.So the numerator is 3+5j.
The denominator is found by breakin up the
√(-16 into √-1 an √16.The same concept applies as above,j^2 is √-1 and √16 is 4.The denuminator nw becomes 8-4j.
The question nw reads 3+5j/8-4j.
We nw hav to find the congugate of the denumenator which is 8+4j.
So this nw reads:
3+5j/8-4jx8+4j/8+4j
(3+5j)(8+4j)
24+12j+40j+20j^2
24+52j+20j^2
24+52j+20(-1)
24+52j-20
24-20+52j
4+52j
The denumenator is
8^2+(4j)^2
64+16j^2
64+16(-1)
64-16
48
1/48(4+52j)
sory the denominator is (3-2j)(3+2j), but the rest of the question suppose to be correct.
ReplyDeletein order to minus the two you can simply minus real with real and imaginary with imaginary
ReplyDelete2 + 3j
-
6 - 4j
_______
-4 + 7j
_______
(2+3j)-(6-4j)
ReplyDeletewhen added a question like this always remeber to add real numbers by themself and imaginery number by themself therefore
real numbers = 2+(-6) = -4
remember a + and a - is a -
imaginery numbers = 3j-(-4j) = 7j
remeber when two minus sign come together we add
so the answer for the question will be
real and imaginery
-4 + 7j
for question no.7
ReplyDelete[3+ √(-25)]*[8-√(-16)]
since the square root of a negative number is the same as the the square root of that same number multiplied by the square root of -1;
we will get: (3+√25√-1)*(8-√16√-1)=
(3+5√-1)*(8-4√-1
now, since √-1=j
we will get: (3+5j)*(8-4j)=
24+40j-12j-20j^2=
24+28j-20j^2=
now, sincej^2=-1
we will get: 24+28j-20(-1)=remember a-*-will result in a +)
hence, 24+28j+20=
44+28j
desi girl
ReplyDeletesorry but i think the denominator is supposed to be: (8-4j)(8+4j)
which will give; 8^2 - 4j^2=
64 - 16(-1)=
64 + 16= 80
any way the rest of the answer is correct, but don't forget the answer isn't complete until this step: 1/80(4+52j)
all the answers looks accurate! keep up the good work guys!!!!!!!!
ReplyDeleteNumber 1
ReplyDelete(2+3j)-(6-4j)
minus real from real and img. from img. so we say
=2-6 +3j-(-4j) minus-minus=+
=-4 + 7j.
Number 2
(2 + 3j)(1 – 5j)
in this case it is multiple so we expand
=(2X1)+(2X-5j)+(3jX1)+(3jX-5j)
= 2 + -10j + 3j + -15j^2
= 2-10j+3j-15j^2 10j-3j=7j
=2-7j-15j^2
j^2 as noted has a value of minus 1 squared
so we replace j^2 with -1
=2-7j-15(-1)
=2-7j+15
=2+15-7j
Ans =17-7j
Number 3
conjugate is defined as the same given value but with the opposite sign
(a) 3+5j would be equal to
= 3-5j
(b) 2-6j would also be equal to
= 2+6j
Number 4
( 3+4j)/(3-2j)
As seen above the denominator is 3-2j therefore the conjugate is 3+2j.
3+4j/3-2j X 3+2j/3+2j
working denominator first.
(3-2j)(3+2j)
= (3x3)+(3x2j)+(-2x3)+(-2jx2j)
= 9 + 6j -6 -4j^2
= 3+6j-4j^2
but remember j^2=-1
= 9 +6j-4(-1)
= 9+6j+4
= 13+6j
= 13+6j
desi girl you j^2= -1 and j= square root -1 and not how you have it as square root -1 = j^2. the rest of the problem is correct. we should always remember that when working with the conjugate the denominator can be seen as (a-b)(a+b)= a^2- b^2 which will work out to give you a real number eliminating the imaginary part.
ReplyDelete[3+√(-25)]+[8-√(-16)]
ReplyDeletenow, 3+√(-25)=
3+√25√-1= (from the known strategies)
3+5j
and, 8-√(-16)=
8-√16√-1=
8-4j
from here on is just simple addition, real with real and imaginary with imaginary...
hence: (3+5j)+(8-4j)=
11+j
@ce: i don't agree with you for number 4
ReplyDeletethis part is correct
(3+4j)/(3-2j)=
[(3+4j)/(3-2j)]*[(3+2j)/(3+2j)]=
when working the denominator first:
it supposed to be: 3^2-2j^2
which is: 9-[4(-1)]=
9+1=
10
next step is to obtain the numerator:
(3+4j)*(3+2j)=expanding the brackets we get: 9+12j+6j+8j^2=
9+18j+(8*-1)=
9+18j-8=
1+18j
the next step is to divide the numerator by the denominator,hence we shall get:
=> 1/10(1+18j)
I hope my response helps you to understand a little more.
no 3
ReplyDeleteto find the cojugate, simply change the opperation sign:
a)3-5j
b)2+6j
for no 5 to 8
ReplyDeletethe first thing to do is to simplify the complex looking terms and then it will be easy to work out
-so it will be 3+5j and 8-4j after we get this all that has to be done from here is the opperations addition, subtraction, multiplication and division.
NB remember for division we have to multiply by the conjugate.
5)(3+ sqrt.-25)+(8- sqrt.-16)
ReplyDelete(3 + sqrt.-1 x sqrt.25)+(8- sqrt. -1 x sqrt.16)
(3 + j x 5) + (8 - j x 4)
(3+5j)+(8-4j)= 11 - 1j
(6)Evaluate [3 + √(-25) ] - [ 8 - √(-16)]
ReplyDeletethe first stem is to work the brackets first
[3 + √25j ] - [ 8 - √i6j]
[3 + 5j ] - [ 8 - 4j]
5+9j
Write the conjugate of
ReplyDeletea) 3 + 5j
b) 2 – 6j
to find the conjugate you just simply have to put the opposite of the operation sign
the conjugate of
(a)3+5j is
3-5j
(b)2-6j
2+6j
Evaluate [3 + √(-25) ] * [ 8 - √(-16)]
ReplyDeletei think that will be;
=(3+5j)*(8-4j)
=24-12j+40j-20j^2
j^2 = -1
=24+20-28j
=44-28j
The real numbers and the imaginary numbers wen added are placed by themselves...meaning we add real numbers by themselves and imaginary numbers by themselves...and the conjugator is basically the negative operator sign..
ReplyDeleteActually the conjugate is just the opposite of the equation not the negative operation sign. Example (3+5) conjugate is (3-5).
ReplyDeleteto solve this question keep in mind that
ReplyDelete√(-25)= √(-1)x √(25) = 5j
√(-16)= √(-1)x √(16) = 4j
now re write
3 + 5j
+
4 - 4j
________
11 + 1j
________
Evaluate [3 + √(-25) ] - [ 8 - √(-16)]
ReplyDeletein order to solve this question follow the same steeps i took in the previous question then minus so :
3 + 5j
-
4 - 4j
________
-1 + 9j
________
Evaluate [3 + √(-25) ] x [ 8 - √(-16)]
ReplyDeletealso to solve this one take the same steeps as i did on the previous one then rewrite and multiply
(3 + 5j)x(4 - 4j)