Let z1 = 2 − 3j, z2 = 4 + 6j.
Evaluate a) z1 + z2 b) z1 - z2 c) z1z2 d) z1/z2 as x + jy, x, y ∈ R.
Find r = z and arg(z) for the following complex numbers
z = 1, j, −1, −j, 1 + j, 1 − j, −1 + j, −1 − j, b) z = 1 + j√3 c) z = −2 + j2√3
and mark them on a drawing of the complex plane.
Find the following powers of j: a) j^8, b) j^42, c) j^11, d) j^105.
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if z1 = 2-3j and z2 = 4+6j then:
ReplyDeletez1+z2=?
you simply substitute for z1 and z2 and you will get (2-3j)+(4+6j)
as i always say real under real imaginary under imaginary.
you will get for the real 2+4=6
and for the imaginary -3j+ (+6j). rememba a plus and a plus gives a plus so you will get
-3j+6j=3j therefore the answer will be 6+3j.
b)Z1-Z2
ReplyDelete=2-3j-4+6j
Put real under real and imaginery under imaginery anyou will get -2+(-9j)
c)Z1Z2
=(2-3j)(4+6j)
Expand brackets
8+12j-12j+18
=26
a)j^8
ReplyDelete= (j^2)(j^2)(j^2)(j^2)
=(-1)(-1)(-1)(-1)
=1
c)if z1 = 2-3j and z2 = 4+6j then;
ReplyDelete(z1) (z2)
(2-3j) (4+6j)
8 + 12j - 12j - 18j^2
since j^2 = -1 simplify and you will get..
8 + 18
26
am i right?
j^11
ReplyDelete= (j^2)(j^2)(j^2)(j^2)(j^2)j
=(-1)(-1)(-1)(-1)(-1)j
= -1j
j^8
ReplyDelete= (j^2)(j^2)(j^2)(j^2)
=(-1)(-1)(-1)(-1)
=1
for 1 part c
ReplyDeleteZ1Z2
simply substitutes to become
=(2-3j)(4+6)
and then you multiply 2 by the 2nd eqn
which wud be
=(2*4)+(2*6j)
then you multiply -3j by the 2nd eqn
which wub be
(-3j*4)+(-3j*6j)
then you add part 1 and 2 together and group the like terms and then solve, which wud be
=(2*4)+(2*6j)+(-3j*4)+(-3j*6j)
=8+12j-12j-18j^2
=8+0j-18j^2
and since j^2 =-1
this means
=8+0j-18(-1)
=26+0j
aha. 1)d) z1/z2 as x + jy, x, y are a set of real nos.
ReplyDeletez1/z2
=(2-3j)/(4+6j)
the conjugate would have to be used as we know, to make the dominator real without affecting the potential value of the expression.
conjugate = 4-6j
z1/z2
=(2-3j)/(4+6j)*(4-6j)/(4-6j)
=(-10-24j)/52
i'll use decimals in this next step because fraction here tend to be a little confusing...for me at least.
= 0.2-0.5j (1dp approx.)
z1/z2
=x+jy
=0.2-0.5j
=0.2+j(-0.5)
values for x and y can be equated:
x=0.2
y=-0.5
j^42
ReplyDelete= (j^2)(j^2)(j^2)(j^2)(j^2)(j^2)(j^2)(j^2)(j^2)(j^2)(j^2)(j^2)(j^2)(j^2)(j^2)(j^2)(j^2)(j^2)(j^2)(j^2)j
= (-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)j
=1j
j^105
ReplyDeletesince j^2=-1
this can be solved by saying
=(-1)^52 (j) and this is equal to
= 1j
can any1 explain to me how to express a term in polar form wit a simple examlple?
ReplyDeletea)add the 1st term in the 1st bracket with the first term in the 2nd bracket, then add the 2nd term in the 1st bracket with the 2nd term in the 2nd bracket. and also beware of police!!!
ReplyDelete=(2-3j) + (4+6j)= 6 + 3j
b)the same thing applies for (b) except here we minus instead of add
=(2-3j) - (4+6j)
= -2 - 9j
c)for problems like this one we simply multiply each term in the 1st bracket by all terms in the 2nd bracket
= 2-3j x 4+6j
= 8 + 12j - 12j - 18j^2
= 26
d)can i get some help with d plzzzzzz
i think i did the question you're asking about higher up mystic. i hope its clear enough.
ReplyDeletebtw, always open to criticism.
i think i'll take a stab at your request Ram. first thing we have to know is that polar simply means dealing with angles. remeber autoCAD?
ReplyDeletewhere an eq'n is in the form: a+bj; the eq'n for polar form: r(cosθ-jsinθ) <== note theta, ie the angle.
let's take 3+4j
on a sketch we would go 3 right on the real axis, then 4 up on the imaginary axis. a line from this point to the origin would give the hypotenuse of a right angle triangle.
we need to find the lenght of this hypotenuse, r by pythagoras theorm:
r = √(a^2+b^2)
in this case:
r = √(3^2+4^2) = √(9+16) = √25 = 5
so far so good. all we need to complete our eq'n in polar form now is θ.
from your sketch you would see that θ, the angle between the hypotenuse and the real axis, is related to your two givens by:
tanθ = opp/adj; therfore,
θ = arctan(opp/adj)
for our example:
θ = arctan(4/3) = 53۫
now we have all we need to form our polar eq'n. r(cosθ+jsinθ) becomes:
5(cos53۫+jsin53۫)
simple enough, right? note that the angle lies in the first quadrant, ie the top right. if it was in another quadrant calculation for θ would be slightly different.
for second quatrant: θ = 180 - calculated angle,
for third quadrant: θ = 270 - calculated angle,
for forth quatrant: θ = 360 - calculated angle,
and from there it's easy pickings.
some things to note though:
1) r may not always be such a nice neat number.
2) the adj side takes the magnitude of the real number and the opp side takes the magnitude of the imaginary number.
3) check the mode of your calculator when working θ. use the DEG mode.
4) pay attention the quadrant of the angle.
i hope this helped. feel free to add to it, or to clarify. thanks for the opportunity Ram.
hey Ram im sorry, i didnt realise you had asked a question. well here's a simple example
ReplyDelete- if you have a complex term such as 4+3j and you want to put this in polar form then the first thing to do is to sketch this in a cartesian plain. the x axis will be real so 4 will be plotted on this axis, while 3 will be plotted on the y axis since this axis is imaginary.
the next step is to form a triangle with the plotted points and lable the angle that is near the origin 'theta'. this angle now has to be found:
lets say theta is x
tan x = opp/adj
tan x = 3/4
x = 37 degrees
NOW THAT WE FOUND THETA, WE HAVE TO FIND MAGNITUDE. THIS IS ACTUALLY FOUND BY PYTHAGORAS THEOREM
= (3^2 + 4^2)^1/2
= 5
so now that we have theta and magnitude we can now put this complex no. in polar form. so it will be
5(cos 37 + j sin 37)
thanks mstic i finally understand this thing
ReplyDeleteif z1 = 2 − 3j, z2 = 4 + 6j.
ReplyDeletethen b) z1 - z2
2 − 3j
-
4 + 6j
________
-2-9j
b) j^42
ReplyDeletesince j^2=-1
then j^42
=-1 or same as 1j
d)j^105
ReplyDeletewell there will be 52 j^2 and one j^1 so
(-1)(-1) 52 times is (+1) and then there is one j so the ans is prob 1j.
I agree wid DJ HALF TIME that the conjugate would have to be used as to make the denominator real widout affecting it..sure rite...
ReplyDeleteKen, i agree that j^42 = -1, but i don't understand why you would say that it's the same as 1j. 1j = 1√-1; -1 = j^2
ReplyDeleteI agree with ken. If you work it consider it properly, 42 is a factor of 2. Hence, it will end up being in its original form. i.e. Its just as saying j^2 = -1
ReplyDeleteI still quite don't understand how to work out j^105 and j^42 in that way can anyone explain further?
ReplyDeleteQuestion 1
ReplyDeletez1+z2=?
Since z1 = 2-3j and z2 = 4+6j then:
you simply substitute for z1 and z2 which is
(2-3j)+(4+6j)
Put real under real and imaginery under imaginery
(2-3j)
+
(4+6j)
=6+3j.
Question
ReplyDelete(√-2)+ j^4