Calculate
(a) (2 − 8j) + (5 + 4j) (b) (2 − 8j) − (5 + 4j) (c) (2 − 8j) × (5 + 4j)
(d) (1 − 7j) × (3 + 10j) (e) (5 − j√2) × (2 + 7j√2) (f) (1 + j)^4
Plot 1 + 3j, 2 − 2j, −4 + j and −2 − 2j on the Argand diagram and for each write as polar coordinates
Calculate the modulus and phase angle of 1 + 3j, 2 − 2j, −4 + j and −2 − 2j.
Solve the quadratic equations
(a) x^2 − x + 1 = 0;
(b) 2x^2 + x + 1 = 0;
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to add (2-8j) to (5+4j) you must remember to put rel under the real and imaginary under the imaginary. for the real you will get 2+5=7 and for the imaginary you will get -8j+(+4j) which is -8j+4j=-4j (remember to look out for the police which is the minus sign). so the answer will be 7-4j, where 7 is the real and -4j is the imaginary.
ReplyDeleteto minus (5+4j) from (2-8j) you must remember to put rel under the real and imaginary under the imaginary. for the real you will get 2-5=-3 and for the imaginary you will get -8j-(+4j) which is -8j-4j=-12j (remember to look out for the police which is the minus sign). so the answer will be -3-12j, where -3 is the real and -12j is the imaginary.
ReplyDeletealways remember that a (minus by a minus= a plus), a(plus by a plus= a plus), and a(minus by a plus OR a plus by a minus= a minus)
ReplyDelete(C) (2 − 8j) × (5 + 4j)
ReplyDeleteto calculate this the 2 must multiply with all the terms in the second bracket,so this will be
2x5+2x4j=10+8j
Then we multiple the -8j with all the terms in the second brackets,so this nw becomes
-8jx5+-8jx4j=-40j-32j^2
DON'T 4GET D POLICE....
SO we nw group al the terms which gives us:
=10+8j-40j-32j^2
=10-32j-32j^2
since j^2 is -1,we replace it,so it nw becomes
=10+8j-40j-32(-1)
=10+8j-40j+32
=10+32+8j-40j (groupin all terms)
=42-32j
excellent work desi girl!!!!!
ReplyDelete(d) (1 − 7j) × (3 + 10j)
ReplyDeletewel remember to expand your brackets, and show your loops so you wont get lost.
wel you take the 1 and multiply it by everything in the second bracket, you will get (1*3+1*10j)=3+10j
you then do the same for the -7j, you will get (-7j*3 + -7j*10j)= -21j+ -70j^2
= -21j-70j^2
but remember j^2=-1
so substituting j^2=-1 in -21j-70j^2
you will get -21j-70(-1)= -21j+70
the final equation will be the first part you worked combined the second part which gives:
3+10j-21j+70
you then group all like terms by rearanging the equation in my case ill group all the real on the left and imaginary on the right which gives;
3+70+10j-21j= 73-11j (the final answer)
Question 1:Add and express in the form of a complex number a + b i.
ReplyDelete(2 + 3i) + (-4 + 5i) - (9 - 3i) / 3
Question 2:Find the complex conjugate to.
ReplyDelete1 + 8i
i am jus giving the answers to the questions as a guide so as to know if ur on the right tract.
ReplyDeletequestion 1: -5 + 9 i
question 2: 1 - 8 i
To calculate the modulus u must square each number and then find the square root of the number.
ReplyDelete1 + 3j
(A) Z=√1^2+ 3^2
=√1+9
=√10
2 − 2j
(B) Z=√2^2+2^2
=√4+4
=√8
To calculate the phase angle its always
tan Ø = imaginery/real where the j is always imaginary.
(a)1 + 3j
tan Ø=3/1
Ø = tan-1 3/1
=71.57
(b)2 − 2j
tan Ø=2/2
tan Ø=1
Ø = tan-1 1
=45.
Calculations
ReplyDeletea) 2 – 8j
+
5 + 4j
= 7 - 4j
b) 2 – 8j
-
5 + 4j
= 7 – 12j
c) ( 2 – 8j) (5 + 4j) = (2 × 5) + (2 × 4j) + (-8j × 5) + (-8j × 4j)
= 10 + 8j + (-40j) + (-32j^2)
= 10 + 8j – 40j – 32j^2
= 10 – 32j – 32j^2
Recall j^2 = -1
This Implies: (-32) × (-1) = 32
So 10 – 32j + 32 = (10 + 32) – 32j
Answer = 42 – 32j
d) ( 1 – 7j) (3 + 10j) = ( 1 × 3)+ (1 × 10j) + ( -7j × 3) + (-7j × 10j)
= 3 + 10j + (-21j) + (-70j^2)
= 3 + 10j – 21j – 70j^2
= 3 -11j – 70j^2
Recall j^2= -1
This implies: 3 – 11j – (- 70)
So 3 – 11j + 70 = (3 + 70) – 11j
Answer = 73 – 11j
e) (5 - j√2) (2 + 7j√2) = (5 × 2) + (5 × 7j√2) - (j√2 × 2) - (j√2 × 7j√2)
= 10 + (35j√2 – 2j√2) – 7j^2 √2√2
Recall √2 × √2 = 2
j^2 = -1
This Implies:
10 + 33j√2 – (7×-1× 2)
= 10 + 33j√2 + 14
= (10 +14) + 33j√2
Answer = 24 + 33j√2
f) (1 + j) ^4
This is the same as:
(1+j) (1+j) (1+j) (1+j)
Now we can simplify (1+j) (1+j)
Recall (a + b) (a + b) = a^2 + 2ab + b^2
This Implies (1+j) (1 + j) = 1 + 2j + j^2
Now we multiply be (1+j) this would now be (1+j) ^3
So, (1 + 2j + j^2) (1 +j)
= (1 × 1) + (1 × 2j) + (1 × j^2) + (1 × j) + (2j × j) + ( j^2 × j)
= 1 + 2j + j^2 + j+ 2j^2 + j^3
Now we multiply be (1 + j) this would make it back to the original state (1 +j)^4
(1+ 2j + j^2 + j + 2j^2 +j^3) (1 + j)
Now we know the first part would be the same because we are multiplying by one first. So we write back the equation. The second part we multiply by j. thirdly we remove the brackets, group like terms and simplify.
Part one and two:
(1 + 2j + j^2 + j + 2j^2 + j^3) + (j + 2j^2 +j^3 +j^2 +2j^3 +j^4)
When all like terms are grouped we get
1 + 4j + 6j^2 + 4j^3 + j^4
Remember j^2 = -1
1 + 4j + 6(-1) + 4(-1) j + (-1)(-1)
= 1 + 4j -6 -4j + 1
Answer = -4 + 0j
Hey angel...
ReplyDeleteI thought that complex nos only dealt wit j's.
those questions u put up are not complex because it doesnt have the j's.
If u recall: √-1 = j and j^2 = -1.... where does "i" come in...
wel dark angel from what i heard and read (i and j) can be used but we normally use (j) especially in electrical as (i) represents current, and to avoid confusion the j is used.
ReplyDeleteoh yea ah get them questions and ans from the internet, but i think i should not have post the ans for the second question i posted as it was very simple.
ReplyDeleteIn the solving of complex numbers we must first remember
ReplyDeletej= √ -1 and j^2 = -1
With Complex Numbers we must always remember:
1. For Addition and Subtraction we must always keep real with real and imaginary with imaginary.
2. An Imaginary Number is the √-ve numbers
3. For Multiplication eliminate brackets by looping or expanding.
4. Division of two complex numbers results in a complex number
5. When multiplying a division problem by its conjugate the denominator MUST work out to be a real number.
in some cases ppl use i dont they?
ReplyDeletei being imaginary
ReplyDeletefor the qudratic equation.....
ReplyDelete(a) x^2 − x + 1 = 0
(x^2 - x)(x + 1)=0
x(x - 1) -1(x-1)
(x - 1)(x - 1)
x - 1=0
x=0
(b) 2x^2 + x + 1 = 0
ReplyDelete2x^2= (2)(-1)
(2x^2+2x)(-x+1)=0
2x(x+1)-1(x+1)
(2x-1)(x+1)
2x-1=0
2x=1
x=1/2
x+1=0
x=-1
(a)(2 − 8j) + (5 + 4j) = 7 - 4j
ReplyDelete(b)(2 − 8j) − (5 + 4j) = -3 -12j
(c)(2 − 8j) × (5 + 4j) = 10 - 32j
(1 − 7j) × (3 + 10j)
ReplyDeletereal is numbers and imaginery numbers are letters and numbers.
3 -21j + 10j -70 j^2
3 -11j -70j^2
3 - 11j - 70 (-1)
-70 (-1)= 70
so u say
3 + 70 - 11j
73 -11j
For(a) (2-8j)+(5+4j) simple addidtion
ReplyDeletewe first put aside the real and imaj.
(2+5) + (-8j+4j)(- + +)=-
which would equal =7 + -4j.
for (b)(2-8j)-(5+4j)
2-5 -8j-4j (-8 - +4)=-12
real imaj.
= -3 - 12J.
for (c) (2-8j)(5+4j)
as i recall when * we expand out ques.
so we say:
(2x5) +(2x4j) +(-8jx5) +(-8jx4j)
=10 + 8j + -40j -32j^2
=10 -32j -32j^2
note:j^2 as learnt in class is = to -1 and we have a j^2 SO REPLACING IN EQ.
=10+8j-40j-32(-1) (-32 x -1 = +32)
=10+8j-40j+32
=10+32+8j-40j
=42-32j
for (d)(1-7j)(3+10j)
like before we use the same process,expand and calculate.
= 1x3 + 1x10j + -7jx3 + -7jx10j.
= 3 + 10j + (-21j)+ (-70j^2)
= 3 -11j -70j^2
we obtained a j^2 again so we say
j^2=-1 now replace it into our 3rd line
3-11j-70(-1) (-70 x -1 = +70)
= 3+70-11j
= 73-11j
In our (e) we deal with the square(shed)and we know we always have to remove the shed.
(5 - j√2) (2 + 7j√2)
=5×2 + 5 x 7j√2 - j√2 x 2 - j√2 x7j√2
= 10 + 35j√2 – 2j√2 – 7j^2 √2√2
as noted in class if you multiple 2 of the same numbers at square root the result will be the number being squared √2 × √2 = 2
and we note agian j^2 = -1
10 + 33j√2 – [(7×-1)× 2]
= 10 + 33j√2 + 14
= 10 +14 + 33j√2
= 24 + 33j√2.
when adding complex numbers ensure that you determine the real from the imaginary....
ReplyDeletea)(2 − 8j) + (5 + 4j)
Firstly group them accordingly. 2+5 and −8j + 4j.
Using the advise given by miss you where you look out for the minus sign aka the police man. This is very cruical, using the calculator add 2 and 5 which gives 7 as the real and −8j and 4j which gives −4j.
The answer is (7 − 4j)
to find the polar cordinates of the following 1 + 3j, 2 − 2j, −4 + j and −2 − 2j first plot it into a graph using ur imaginary numbers on your y axis and your real numbers on your x axis. a right angle will be formed between the slope and the x axis. your imaginary number will be your opposite and your real nuumber will be your adjecent. negative signs will be left out because it is to show direction. so.:
ReplyDeletefor 1 + 3j r = √3^2+1^2 = √10 = 3.61
tan Ѳ = opp/adj = 3/1
Ѳ = tan^-1 3 =71.56
for 2 − 2j r = √2^2+2^2 = √8 = 2.83
tan Ѳ = opp/adj = 2/2
Ѳ = tan^-1 1 = 49
for −4 + j r = √1^2+4^2 = √17 = 4.1
tan Ѳ = opp/adj = 1/4
Ѳ = tan^-1 0.25 = 14
for -2 − 2j r = √2^2+2^2 = √8 = 2.83
tan Ѳ = opp/adj = 2/2
Ѳ = tan^-1 1 = 49
1a)2-8j+5+4j=2+5-8j+4j=7-4j
ReplyDeletewell I noticed that there is something wrong with computation calculations for a) and b) because you could never use that approach to solve for x because right now if he/she simplified back that factorized part of the equation it would not work out back to the question giving............you would have to use the quadractic formula to solve for x.......
ReplyDeletefirstly j means imaginary, in complex numbers there is a real and imaginary, real is added to real and imaginary to imaginary.
ReplyDelete(2-8j)+(5+4j)= 2+5(real) (-8j+4j)imaginary
7-4j
Part a: (2-8j)+(5+4j)
ReplyDeleteStep 1: Put all real numbers under each other,then put all imaginary numbers under each other.
Step 2:Then add the real numbers.
Step 2:Add all the imaginary numbers(look out for a policeman/a negative sign).
The result:
2 - 8j Note:Watch out for the
+ police man
5 + 4j
--------
7 - 4j
--------
Part b:
Step 1:Pot all real numbers under each other,and put all imaginary numbers under each other.
Step 2: Take away/- the second real number from the first real number.
Step 3: take away/- the second imaginary number from the first imaginary number.
Note: Watch out for negative signs/police man.
The result:
2 + 8j
-
5 + 4j
________
-2 - 12
- - -----
(a) (2 − 8j) + (5 + 4j)
ReplyDeleteThe numbers are called real; the terms with j are called imaginary. when adding complex numbers, the real numbers are added together and the imaginary are terms are added together
(2 − 8j) + (5 + 4j) =
2 – 8j +
5 + 4j
7 – 4j
(c) (2 − 8j) × (5 + 4j)
This is done like any other expansion of brackets. The first number is multiply by the terms in the second brackets and the second term of the first bracket is multiplied by the terms in the second brackets
(2 − 8j) × (5 + 4j) =
{2 (5 + 4j)} – {8j (5+ 4j)}
= (2*5 + 2*4j) – (8j*5 - 8j * 4j)
= 10 + 8j – 40j-32j^2
Recall: j^2 = -1
Therefore 32j^2 = -32 * -1= 32
10 + 8j – 40j + 32
= 42 – 32j
(f) (1 + j) ^4
1^4 = 1 * 1 * 1 * 1 = 1
J^4= (j^2)^2 (recall J^2 = -1) = -1* -1 = 1
(1 + j) ^4 = 1 + 1 = 4
2-8j +5+4j=2+5 -8j+4j=7 -4j
ReplyDeleteb)2-8j -5+4j=2-5 -8j-4j=-3 -12j
ReplyDeletec)2-8j x 5+4j=
ReplyDelete2x5+2x-4j -8jx5-8jx4j=10-8j -40j-34j^2=42-32j
d)1-7jx3+10j=1x3+1x10j -7jx3-7jx-10j=3+70+10j-21j=73-11j
ReplyDelete5-jsqrt2x2+7jsqrt2=10+35jsqrt2 -2jsqrt2-7j^2=1o+33j+2+7x2=26+33j
ReplyDelete(1+j)^4=1^2+j^2+2x1j=1+2j+j^2(same as before) =1+2j-j^2=4j^2=-4
ReplyDeleteHey guys, here are some simple questions on Complex Numbers to try:
ReplyDeleteConvert from rectangular form to polar form.
• 4 + j^6
• 23 + j^17
• -8.1 + j^6
Here is an example to start you off:
Express (7 + j6) in polar form.
First determine your x and y values (i.e. the real part and the imaginary part).
x = 7 and y = 6
Secondly determine the resultant,r: z = x + jy
Now, notice is you draw z on a graph, you can see that r can be obtained by using Pythagoras’s thermo;
r = √(x^2 + y^2) = √(72 + 62)
r = √85
And, θ = tan-1(6/7) = 40.60o
Now,
Say we were only given θ & r and we were to find y?
sinθ = opp/hyp = y/r
y = r sinθ ----------- (imaginary = j)
Similarly,
If we had to find r, the same principle applies.
cosθ = adj/hyp = x/r
x = r cosθ ------------- (real = r)
Hence,
z can be represented in polar form by:
z = √85 cos(40.60o) + j sin(40.60o)
first step i take in doing complex numbers i look for real and imaginary numbers find them and give them partners if i can no one likes to be lonely .i cant do this with out a calculator take my advices does police men will catch you i make a lot of mistakes if i don't use my calculator i hate police
ReplyDeletewhen adding or subtracting complex number it easier to do it this way
ReplyDelete(a) (2 − 8j) + (5 + 4j
real imaginary
( 2 − 8j)
+(5 + 4j
= 7 - -4j
(b) (2 − 8j) − (5 + 4j)
ReplyDeleteusing the same method show above
(b)(2 − 8j) − (5 + 4j) = -3 -12j
multiplying complex numbers the first step is to remove the brackets and then expanding the terms then solve
ReplyDeletec) ( 2 – 8j) (5 + 4j) =
(2 × 5) + (2 × 4j) + (-8j × 5) + (-8j × 4j)
= 10 + 8j + (-40j) + (-32j^2)
= 10 + 8j – 40j – 32j^2
= 10 – 32j – 32j^2
j^2= -1
-32x-1=+32
there for answer= 42 – 32j
a) 2-8j + b)2 − 8j − 5 + 4j
ReplyDelete5+4j = -3-12j
= 7+4j
1+3j
ReplyDeletewe find modulus using pythagoras theorem
= (1^2 + 3^2)^1/2
= 3
the phase angle is:
tan x = opp/adj
tan x = 3/1
= 72 degrees
2-2j
modulus-
=(2^2 + (-2^2) )^1/2
= 2.8
the phase angle is:
tan x = opp/adj
tan x = -2/2
x = -45
since the phase angle is in the forth quadrant then it will be:
360-(-45) = 405 degrees
-4+j
the magnitude is:
= (-4^2 + 1^2)^1/2
= 4.1
the phase angle is:
tan x = 1/-4
x = -14
since the phase angle is in the second quadrant it will be:
180 - (-14)= 194 degrees
-2-2j
the modulus is:
=(-2^2 + -2^2)^1/2
= 2.8
the phase angle is:
tan x = -2/-2
x = 45
since the phase angle is in the third quadrant it will be:
x = 180 + 45
x = 225 degrees
(4+3j)divided by (5-6j)
ReplyDeletecan someone explain the steps in modulus in details cause i think desigirl is saying something different from mysticwings
ReplyDeleteTo Calculate the modulus and phase angle of
ReplyDelete1 + 3j
the first step is to calculate the modulus (magnitude) this is done by using
Pythagoras theorem
= (1^2 + 3^2)=10
square root of 10=
modulus= 3.2
the phase angle could be calculated by sketching a graph and finding the angle θ
tan θ = opp/adj
tan θ = 3/1
θ = tan-1 3
= 72 degrees
wait can some one plz explain to me what phase angle is all about
ReplyDeletewell animechick the phase angle is best found by plotting the graph.For example if you have 3=5j on a graph 3 would be on the x axis and 5 would be on the y axis.You just plot the (3,5)point and connect it to the origin and from that line as the hypotneuse form a right angle triangle with the x axis. The phase angle is between the x axis and the hypotenuse checking from an anticlockwise direction,like if the triangle is formed in the lower right quadrant the pahse angle would be 360-the angle in the triangle.
ReplyDeletei dont know if that was the answer you are looking for
a) (2 − 8j) + (5 + 4j)
ReplyDelete= 7 - 4j
(b) (2 − 8j) − (5 + 4j)
= -3 - 12j
(c) (2 − 8j) × (5 + 4j)
expand the brackets
= 10 - 40j + 8j - 32j squared
j squared = -1
therefore
= 10 - 32j - 32
For addition and subtraction of complex u keep real values under real and imaginary under imaginary ...for multiplication u get rid of brackets by simply looping as miss showed us..or expanding..very easy and for division two complex gives a complex number..
ReplyDeleteAnd the division between a real and an imaginery number gives a conjugate. Correct me if am wrong
ReplyDeleteit doesn't give a conjugate. you use the conjugate to make the division of two complex numbers possible.
ReplyDeleteyou can't really divide a real number by an imaginary one; that's why the conjugate is needed. on the other hand, you can divide an imaginary number by a real one.
I agree division between real and imaginery does not give you a conjugate, but instead you have to multiply by the conjugate, you at least had an idea.
ReplyDeleteA)(2 − 8j) + (5 + 4j)
ReplyDeleteAdd real numbers with real number and imaginary with imaginary
= 5 + 4j
B)(2 − 8j) − (5 + 4j)
Add real numbers with real number and imaginary with imaginary and remember your signs when a - sign and a + sign come together its a -
= 3 - 12j
C)(2 − 8j) × (5 + 4j)
Multiply real numbers with real number and imaginary with imaginary
= 10 - 32j
D)(1 − 7j) × (3 + 10j)
Multiply real numbers with real number and imaginary with imaginary
= 3 - 70j
Thanks for the correction..
ReplyDelete