Evaluate a) z1 + z2 b) z1 - z2 c) z1z2 d) z1/z2 as x + jy, x, y ∈ R.
Find the modulus and arg(z) for the following complex numbers
- 25
- 6j
- −9,
- −12j
- 11 +12 j
- 41 −6 j
- −16 + 8j
- −32 −76 j
c) z = −7 + 6j and mark them on a drawing of the complex plane.
Evaluate a) j^18, b) j^72, c) j^31, d) j^505.
Complex numbers involves various calculations..addition, subtraction, multiplication and division which can b tricky lol..for the fist question part (a) this is probably the first calculation we've learnt on complex numbers...addition...its quite simple i think...jus as long as you do what we've learnt...write down the equivalents of z1 and z2...make sure you put all the 'real' in one section and the 'imaginary' in the other column and add them...beware of
ReplyDeleteR I
the 'police' lol so we have z1--> 2 -5j +
z2--> 3 +6j
the answer when we add both is '5 + j'
well i don't think we are allowed to answer more than one question but i guess its ok to give ideas right?...so for part (b) the same steps mentioned above can be followed...its just to subtract instead of adding...in this case however, you must especially beware of the sign changes or "police" as miss calls it lol...for part (c) this involves multiplication so its easier when you put both terms in brackets and multiply each term by both terms of the other bracket and for part (d) i think division can be tricky somtimes but as long as you remember to multiply the whole expression by the conjugate of the denominator which in this case is '3 + 6j' so the conjugate is "3 - 6j"...so i hope these lil tips help...we all learn from each other so if i made any errors please let me kno..thanks
ReplyDeleteZ1 + Z2
ReplyDelete(2-5j) + (3+6j)
(5+j)
z1 - z2
(2 - 5j) - (3 +6j)
( -1-11j)
b)z= 5 +j√3
= 5 + √3 (-1)
j^18
= (-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)
= -1
since z1= 2-5j and z2=3+6j,
ReplyDeletez1+z2= 5+j
z1-z2= -1-11j
z1z2= (2-5j)(3+6j)= 36-3j
When dividing complex numbers you include all the fundamental steps of maths which is if ppl do nt knw is addition, subtraction, division and multiplication.
ReplyDeletez1 = 2 − 5j , z2 = 3 + 6j
z1/z2 = 2 − 5j / 3 + 6j
what you do from here is that you multiply by the conjugate of the denominator which is the one on the bottom. The conjugate involves just reversing the sign of the complex number.
z1/z2 = (2 − 5j / 3 + 6j) x (3 − 6j / 3 − 6j)
Step 1. multiply the numerators which are the complex numbers on top.
(2 − 5j x 3 − 6j) = (6 − 12j − 15j + 30j^2)
remember from the basics in complex no. that j^2 is equal to −1.
(6 − 12j − 15j + (−1)30) = (6 − 12j − 15j −30)
Step 2. add like terms, the real numbers to the real no. and the complex to each other.
(6 − 12j − 15j − 30) = (−24 − 27j)
Step 3. multiple the denominator, does this look familiar. Yes it does it is a difference of two squares. If you do see it then look at it this way......
(a2 − b2)(a2 + b2) = a2 + ab − ab − b2)
= a2 − b2
(3 − 6j x 3 − 6j) = (9 + 36j^2)
= (9 − 36)
= −27
Now put the ans of step 1. / step 3.
z1/z2 = (−24 − 27j)/−27
Q1 part d: z1/z2 as x +jy,x, y 'the funny looking e'R; wat does this mean?
ReplyDeletefor the last question:
ReplyDeletej^18 = -1
j^72 = 1
j^31 = -1j
j^505 = 1j
an easy way to work out problems like these is to divide the power by 2 and if it is an even number then the answer is 1, but if it is odd number then the answer is -1.
for the first
ReplyDeletej^18 = -1
could it be worked like as follows
-1 = j^2
therefore j^18 = j^2
right?
then bring all the like terms(meaning the j's of course) onto one side and get
j^18 divided by j^2 = 0
and get the result of
j^(18-2)=0
j^(16)=0
IF as misticwings* says to divide the power by 2 and if it is an even number then the answer is 1, but if it is odd number then the answer is -1.then the overall result will be 0
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
if z1 = 2 − 5j, z2 = 3 + 6j
ReplyDeletethen z1xz2=
(2 − 5j) x (3 + 6j)
6+18j-15j-30j^2
36+3j
yeah u're correct.Ken is the method u're using the one that we use for impendence
ReplyDeletewhen multiplyin complex numbers, both the numerator and the denominator ar mulpiplied b the conjugate denominator(the opposite sign)
ReplyDeletez1 = 2 − 5j, z2 = 3 + 6j.
d) z1/z2
= 2 − 5j/3 + 6j
= (2-5j)(3-6j)/(3+6j)(3-6j)
numerator: 2(3-6j)- 5j(3-6j)
= 6-12j - 15j + 30j^2 (j^2=-1)
= -24-27j
denominator:(a+b)(a-b)=a^2-b^2
= 9-36j^2 (j^2=-1)
= 9+36=45
= -24-27j/45
= (-24/45)- 27j
i dont understand x + jy, x, y ∈ R.
Can someone tell me what is the modulus of agar? Am a bit confused with how the question is phrased....................
ReplyDeleteThe modulus of agrand diagram is simply asking for the magnitude which is r
ReplyDeleteOhh. thanks alot darky
ReplyDeleteQuestion:
ReplyDeleteLet z1 = 2 − 5j, z2 = 3 + 6j.
Evaluate a) z1 + z2 b) z1 - z2 c) z1z2 d) z1/z2 as x + jy, x, y ∈ R.
Recall; when adding or subtracting complex numbers, you add or subtract the corresponding quantities (i.e. the real with the real and the imaginary with the imaginary)
Answer:
a) Z1 + Z2 = (2 - 5j) + (3 + 6j)
= 2 + 3 - 5j + 6j
= 5 + j
Answer:
ReplyDeleteb) Z1 - Z2 = (2 - 5j) + (3 + 6j)
= 2 - 3 -5j - 6j
= -1 - 11j
Recall; when dividing complex numbers you mutilpy by its complex conjugate. This means whatever the operation (+ve or -ve) in the denominator you multiply the complex number (both numerator and denominator) by this denominator, but first you must change the original operation (i.e. from +ve to -ve and vice versa).
ReplyDeleteNOTE: that doing this will not affect the value of the complex number because multiply both numerator and denominator by the same quantity is the as multiplying it by 1.
E.g. (5 x 9)/ (6 x 9) = (5/6) x 1 = 5/6
Answer:
d) Z1/Z2 = (2 - 5j)/(3 + 6j)
= [(2 - 5j)/(3 + 6j)] x [(3 - 6j)/(3 - 6j)]
= [(2 - 5j)(3 - 6j)] / [(3 + 6j)(3 - 6j)]
Note that the denominator is the difference of two squares;
i.e. (a^2 + b^2)(a^2 - b^2) = a^2 - b^2
Hence the equation will now be:
Z1/Z2 = [(2 - 5j)(3 - 6j)] / (3^2 - {6j}^2)
Now; expanding the numerator,i.e.
2x3 - 2x6j - 5x3j + 5jx6j = 6 - 27j + 30j^2
but j^2 = -1;
which imply: 6 - 27j + 30(-1) = -27j - 24
Thus we have:
(-27j - 24) / (3^2 - 36(-1))= (-27j - 24) / 45
Let z1 = 2 − 5j, z2 = 3 + 6j.
ReplyDeleteEvaluate a) z1 + z2 b) z1 - z2 c) z1z2 d) z1/z2 as x + jy, x, y ∈ R.
Answer:
a) Z1 + Z2 = (2 - 5j) + (3 + 6j)
= 2 + 3 - 5j + 6j
= 5 + j
b) Z1 - Z2 = (2 - 5j) + (3 + 6j)
= 2 - 3 -5j - 6j
= -1 - 11j
Answer:
d) Z1/Z2 = (2 - 5j)/(3 + 6j)
= [(2 - 5j)/(3 + 6j)] x [(3 - 6j)/(3 - 6j)]
= [(2 - 5j)(3 - 6j)] / [(3 + 6j)(3 - 6j)]
Note that the denominator is the difference of two squares;
i.e. (a^2 + b^2)(a^2 - b^2) = a^2 - b^2
Hence the equation will now be:
Z1/Z2 = [(2 - 5j)(3 - 6j)] / (3^2 - {6j}^2)
Now; expanding the numerator,i.e.
2x3 - 2x6j - 5x3j + 5jx6j = 6 - 27j + 30j^2
but j^2 = -1;
which imply: 6 - 27j + 30(-1) = -27j - 24
Thus ,
(-27j - 24) / (3^2 - 36(-1))= (-27j - 24) / 45
11 +12 j
ReplyDeleteTo find the modulus of the above set of complex number you just use phytagoras theorm to find the modulus
root of 11^2 + 12^2
Z1Z2
ReplyDelete(2 - 5j)(3 + 6j)
expand the brackets
6 - 15j + 12j - 30j^2
6 - 3j - 30(-1) [j^2 = -1]
6 -3j +30
36 - 3j