untouchablecuz
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not 100% sure how to proceed
revised edition
hey lukybear, ill answer you later (sorry), i've got to go now
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Imaginay Nos (1 Viewer)
- Thread starter Lukybear
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revised edition
hey lukybear, ill answer you later (sorry), i've got to go now
actually i think you are right but you just got confused
The locus of 1/(z-3) was a circle of same centre so 1/(z-3) + 17/3 equals to a locus of (x-17/3)^2+y^2=1/9. (51/9=17/3)
This is because every point 1/(z-3) on the circle shifts right by 17/3, so logically the centre of the circle should shift the same amount
Alternate method
|z-3|=3
|1/(z-3)|=1/3
Thus |Z-17/3|=1/3 and hence the locus of Q is a circle with centre 17/3 and 1/3
I do feel very glad that I am right for the first time. But could you please expand on that?
|z-3|=3 has centre of (3,0) no? not (0,0) If it was shifted by 17/3, it would be z- (17/3 +3)? I am slow to get things so...
1st chapter.
I think the one i posted was it? Is it in the supplement book?
untouchablecuz
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g-g-g-g-g-g-g-geniusssssssssssssssssss
untouchablecuz
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polynomial division should be in Polynomials
it works in exactly the same way as division with real factors
e.g.
divide by x-3
divide by x-(1+i)
letting x-(1+i)=(x-a) and THEN dividing may be of help to you conceptually
K
khorne
Guest
As stated:
|z-3| = 3
|1/z-3| = 1/3
|Z| = |1/z-3| + 17/3
i.e |Z - 17/3| = 1/3
Additionally, while you may already know, these problems are no longer in the syllabus.
O reali? Thats fantastic. Ive been having trouble with them for so long...
O thxs Khorne. I do get it now.
Anyhow, it is good practice tho.
K
khorne
Guest
Actually, according to the syllabus itself:
The student is able to:
• given equations Re(z) = c, Im(z) = k (c, k real), sketch lines parallel to the
appropriate axis
• given an equation | z – z1 | = | z – z2 |, sketch the corresponding line
• given equations | z | = R, | z – z1 | = R, sketch the corresponding circles
• given equations arg z = q, arg(z – z1) = q, sketch the corresponding rays
• sketch regions associated with any of the above curves (eg the region
corresponding to those z satisfying the inequality (| z – z1 | ² R)
• give a geometrical description of any such curves or regions
• sketch and describe geometrically the intersection and/or union of such
regions
• sketch and give a geometrical description of other simple curves and regions.
Topic 2: Complex Numbers
Applications, Implications and Considerations
• Typical curves and regions are those defined by simple equations or inequalities,
such as
Im(z) = 4, | z – 2 – 3i | = | z – i |, | z – 3 + 4i | = 5, Re(z) > 2,
0 < arg z < ¹/2, 0 ² Im(z) ² 4.
• Simple intersections, such as the region common to | z | = 1 and
0 ² arg z ² ¹/4, and corresponding unions, need be done.
• Examples need only involve replacing z by z = x + iy in relations such as
2 | z | = z + + 4, z + > 0, | z2 – ( )2| < 4. They need not include discussion of
curves such as w = z-i/z+i, where z lies on a unit circle.
So you see, it's pretty simple stuff.
z i
-
+
z z z
31
The student is able to:
• given equations Re(z) = c, Im(z) = k (c, k real), sketch lines parallel to the
appropriate axis
• given an equation | z – z1 | = | z – z2 |, sketch the corresponding line
• given equations | z | = R, | z – z1 | = R, sketch the corresponding circles
• given equations arg z = q, arg(z – z1) = q, sketch the corresponding rays
• sketch regions associated with any of the above curves (eg the region
corresponding to those z satisfying the inequality (| z – z1 | ² R)
• give a geometrical description of any such curves or regions
• sketch and describe geometrically the intersection and/or union of such
regions
• sketch and give a geometrical description of other simple curves and regions.
Topic 2: Complex Numbers
Applications, Implications and Considerations
• Typical curves and regions are those defined by simple equations or inequalities,
such as
Im(z) = 4, | z – 2 – 3i | = | z – i |, | z – 3 + 4i | = 5, Re(z) > 2,
0 < arg z < ¹/2, 0 ² Im(z) ² 4.
• Simple intersections, such as the region common to | z | = 1 and
0 ² arg z ² ¹/4, and corresponding unions, need be done.
• Examples need only involve replacing z by z = x + iy in relations such as
2 | z | = z + + 4, z + > 0, | z2 – ( )2| < 4. They need not include discussion of
curves such as w = z-i/z+i, where z lies on a unit circle.
So you see, it's pretty simple stuff.
z i
-
+
z z z
31
Um not quite right
Z=1/(z-3)+17/3 does not imply |Z|=|1/(z-3)|+17/3.... think triangle inequality...
rather Z-17/3=1/(z-3) --->|Z-17/3|=1/3
Explanation:
Instead of considering z, consider the complex number m=1/(z-3).
|z-3|=3 implies |1/(z-3)|=1/3, which implies |m|=1/3, so it means for any given z on the circle, the point m will lie on another circle centred at the origin with radius 1/3. Now noting Z=1/(z-3)+17/3, or Z=m+17/3, which means for any m on the circle stated above, Z is that point shifted 17/3 to the right. So every point on the circle centred at the origin with radius 1/3 is shifted to the right by 17/3 to obtain Q, so logically the centre of that circle should be shifted to the right by 17/3 to obtain the centre for the locus of Q.
Yeah... but some one told me complex is a tool for later chapters. But i do know your absoulutely rite and thxs for warning me.
ninetypercent
ninety ninety ninety
can i assume that you mean cos(pi/5) - cos(2pi/5) = 0.5??
note that
from sum of roots
because
note that
because
ninetypercent
ninety ninety ninety
for second one,
note that
using product of roots
sin(3pi/10) = -sin(17pi/10)
which appears to be true on my calculator
note that
using product of roots
sin(3pi/10) = -sin(17pi/10)
which appears to be true on my calculator
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