questions for the smart ppl (mainly complex nos some graphs and conics) (1 Viewer)
- Thread starter .ben
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could you expand a bit on this? thanks!Stan.. said:The conics question is simple, just sub in either the X or Y value and solve the Quadratic.
but it turns out to be a really messy surd and i don't know if its right.
The differentiation, Do you know implicit differentiation? yeh but still can't get the desire result
The graphs, that just means that the graph is half as wide as before.
P
pLuvia
Guest
Graphs
y=f(x)
y=f(2x)
For the second graph you would contract the first graph by 2 in the x-direction i.e.
For
f(x) | f(-2) | f(-1) | f(1) | f(2)
f(2x)| f(-4) | f(-2) | f(2) | f(4)
I'll give you a picture once I have drawn it
Complex Numbers
1a). It is unit circle with centre (0,2)
b)
2. The locus of z is a perpendicular bisector
|z-2+i| = |z+1+3i|
(x-2)2 + (y+1)2 = (x+1)2 + (y+3)2
Expand and simplify and you should get
6x + 4y + 5=0
3. arg(z-i) / (z+i) = pi/4
Is also equal to
arg (z-i) - arg (z+i) = pi/4
Hence the locus is a major arc of a circle to the left of the y axis
4.
5. a = -1+i
(i) a = sqrt 2 cis 3pi/4
(ii)
z4 = -4
= 4 cis 0
z = 41/4 cis (2kpi / 4) (where k = 0, +1, 2)
When k = 0 z = sqrt2 cis 0
When k = 1 z = sqrt2 cis pi/2
When k = -1 z = sqrt2 cis -pi/2
When k = 2 z = sqrt2 cis pi
Hm.. seems to be something wrong here This is as far as I can go for now
y=f(x)
y=f(2x)
For the second graph you would contract the first graph by 2 in the x-direction i.e.
For
f(x) | f(-2) | f(-1) | f(1) | f(2)
f(2x)| f(-4) | f(-2) | f(2) | f(4)
I'll give you a picture once I have drawn it
Complex Numbers
1a). It is unit circle with centre (0,2)
b)
2. The locus of z is a perpendicular bisector
|z-2+i| = |z+1+3i|
(x-2)2 + (y+1)2 = (x+1)2 + (y+3)2
Expand and simplify and you should get
6x + 4y + 5=0
3. arg(z-i) / (z+i) = pi/4
Is also equal to
arg (z-i) - arg (z+i) = pi/4
Hence the locus is a major arc of a circle to the left of the y axis
4.
5. a = -1+i
(i) a = sqrt 2 cis 3pi/4
(ii)
z4 = -4
= 4 cis 0
z = 41/4 cis (2kpi / 4) (where k = 0, +1, 2)
When k = 0 z = sqrt2 cis 0
When k = 1 z = sqrt2 cis pi/2
When k = -1 z = sqrt2 cis -pi/2
When k = 2 z = sqrt2 cis pi
Hm.. seems to be something wrong here
This is as far as I can go for now
Last edited by a moderator:
Ok, For the first question..ben said:
E: x^2/16 + y^2/9 = 1
Tangents are drawn from P(3, 4)
If you substitute x=3, you will get two answers. Same for y, Find the gradient and then the first questions done. (Two points is probably the only way) Didn't come out either.
The second,
Didn't come out for me.
The third,
Pluvia looks like he wants to answer that. So I'll leave it to him to explain.
Last edited:
YBK
w00t! custom status!! :D
and radius 1pLuvia said:
P
pLuvia
Guest
That's why I called it a Unit Circle But it should clarify what it is to people who don't know
But it should clarify what it is to people who don't know YBK
w00t! custom status!! :D
Ohhh... so that's what a unit circle ispLuvia said:That's why I called it a Unit Circle
I thought that a unit circle was just a normal circle.. lol
Trev
stix
11.
[√(3+4i)+√(3-4i)]<sup>2</sup>
=3+4i+2√(3<sup>2</sup>-(4i)<sup>2</sup>)+3-4i
=6+2√25
=16
[√(3+4i)+√(3-4i)]<sup>2</sup>
=3+4i+2√(3<sup>2</sup>-(4i)<sup>2</sup>)+3-4i
=6+2√25
=16
Trev
stix
8.
I think you're supposed to find coordinates of C.
I would move A to position (0,0) by adding -1-i to the complex coordinates.
Doing the same to point B gives it's new complex coordinates as 4+3i.
[This is just like moving the line around, I forget what the term for it is I think it's translation?]
Since it is a right angle at point A then C would be where you multiply B by i and i<up>3</sup>.
Coordinates of (the translated? point C) is (4+3i)*i and (4+3i)*i<sup>3</sup>; -3+4i and 3-4i.
Then to translate back to original position add 1+i.
Coordinates of C are therefore -2+5i and 4-3i.
I think you're supposed to find coordinates of C.
I would move A to position (0,0) by adding -1-i to the complex coordinates.
Doing the same to point B gives it's new complex coordinates as 4+3i.
[This is just like moving the line around, I forget what the term for it is
I think it's translation?]Since it is a right angle at point A then C would be where you multiply B by i and i<up>3</sup>.
Coordinates of (the translated? point C) is (4+3i)*i and (4+3i)*i<sup>3</sup>; -3+4i and 3-4i.
Then to translate back to original position add 1+i.
Coordinates of C are therefore -2+5i and 4-3i.
Trev
stix
10.
arg(z-2i)-arg(z-2)=pi/2.
Hopefully the picture explains...
arg(z-2i)-arg(z-2)=pi/2.
Hopefully the picture explains...
There was a good reason why 1(b) was left out lol. It's not as easy as it looks. There may be an easier way to do 1(b) but this was what I thought of:
From the diagram, we want to find where the line y=mx is a tangent to x+(y-2)2=1.
Substituting y=mx into the equation of circle and rearranging, we obtain:
(m2+1)x2-4mx+3=0
Now ∆=4m2-12=0, since we want to find the values of m that make it a tangent, hence the 1 intersection point.
.'.m=sqrt3
Now sub. m=sqrt3 into the equation of circle, expanding and simplify to obtain the quadratic
4x2-(sqrt12)x+3=0
Solving this yields x=1/2
.'. y=(sqrt3)/2
Now we can find arg(z) by tan@=sqrt3, where @ is the angle in the postive direction formed by y=(sqrt3)x.
.'. arg(z)=pi/3 or 60o
From the diagram, we want to find where the line y=mx is a tangent to x+(y-2)2=1.
Substituting y=mx into the equation of circle and rearranging, we obtain:
(m2+1)x2-4mx+3=0
Now ∆=4m2-12=0, since we want to find the values of m that make it a tangent, hence the 1 intersection point.
.'.m=sqrt3
Now sub. m=sqrt3 into the equation of circle, expanding and simplify to obtain the quadratic
4x2-(sqrt12)x+3=0
Solving this yields x=1/2
.'. y=(sqrt3)/2
Now we can find arg(z) by tan@=sqrt3, where @ is the angle in the postive direction formed by y=(sqrt3)x.
.'. arg(z)=pi/3 or 60o
P
pLuvia
Guest
11
arg(z-2i) - arg(z-1) = pi/2
To put it this way and an easy way
Let arg(z-2i) = A
Let arg(z-1) = B
.: A - B = pi/2 { Hence A>B }
The locus of z is a circle with centre (1/2, i), and radius = 5/4
arg(z-2i) - arg(z-1) = pi/2
To put it this way and an easy way
Let arg(z-2i) = A
Let arg(z-1) = B
.: A - B = pi/2 { Hence A>B }
The locus of z is a circle with centre (1/2, i), and radius = 5/4