9 Complex/Ellipse Questions help please ASAP!!! (1 Viewer)

Accuracy

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Could someone please help me with the following questions PLEASE (preferably fully worked solutions) . MY HALF YEARLIES ARE IN 2 DAYS );

1) untitled.JPG part (i) , (iii) [solved]
2) untitled.JPG part (ii) [solved]
3) untitled.JPG part (iv)
4) untitled.JPG part (ii)
5) untitled.JPG both parts
6) untitled.JPG
7) untitled.JPG
8) untitled.JPG all parts
9) untitled.JPG all parts
 
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deswa1

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1) i)Cuts the x axis when y=0. Therefore at -2,0,2.
iii) Using product rule+chain rule
<a href="http://www.codecogs.com/eqnedit.php?latex=y=x\sqrt{4-x^2} \\\frac{dy}{dx}=\sqrt{4-x^2}-\frac{x^2}{\sqrt{4-x^2}}\\ \frac{dy}{dx}=\frac{4-2x^2}{\sqrt{4-x^2}}\\ \frac{dy}{dx}=\frac{2(2-x^2)}{\sqrt{4-x^2}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y=x\sqrt{4-x^2} \\\frac{dy}{dx}=\sqrt{4-x^2}-\frac{x^2}{\sqrt{4-x^2}}\\ \frac{dy}{dx}=\frac{4-2x^2}{\sqrt{4-x^2}}\\ \frac{dy}{dx}=\frac{2(2-x^2)}{\sqrt{4-x^2}}" title="y=x\sqrt{4-x^2} \\\frac{dy}{dx}=\sqrt{4-x^2}-\frac{x^2}{\sqrt{4-x^2}}\\ \frac{dy}{dx}=\frac{4-2x^2}{\sqrt{4-x^2}}\\ \frac{dy}{dx}=\frac{2(2-x^2)}{\sqrt{4-x^2}}" /></a>
 

Accuracy

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1) i)Cuts the x axis when y=0. Therefore at -2,0,2.
iii) Using product rule+chain rule
<a href="http://www.codecogs.com/eqnedit.php?latex=y=x\sqrt{4-x^2} \\\frac{dy}{dx}=\sqrt{4-x^2}-\frac{x^2}{\sqrt{4-x^2}}\\ \frac{dy}{dx}=\frac{4-2x^2}{\sqrt{4-x^2}}\\ \frac{dy}{dx}=\frac{2(2-x^2)}{\sqrt{4-x^2}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y=x\sqrt{4-x^2} \\\frac{dy}{dx}=\sqrt{4-x^2}-\frac{x^2}{\sqrt{4-x^2}}\\ \frac{dy}{dx}=\frac{4-2x^2}{\sqrt{4-x^2}}\\ \frac{dy}{dx}=\frac{2(2-x^2)}{\sqrt{4-x^2}}" title="y=x\sqrt{4-x^2} \\\frac{dy}{dx}=\sqrt{4-x^2}-\frac{x^2}{\sqrt{4-x^2}}\\ \frac{dy}{dx}=\frac{4-2x^2}{\sqrt{4-x^2}}\\ \frac{dy}{dx}=\frac{2(2-x^2)}{\sqrt{4-x^2}}" /></a>
hey, the answers for 1) only have -2 and 2. theres no zero for some reason :S
 

Carrotsticks

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Question 1

(i) Let y=0 and solve. x=0 (double root) and x= plus/minus 2

(iii) Differentiate it using the product rule.

Question 2

(ii) z = sqrt(2) cis (pi/12). We want it to be real (this will first occur when the argument is pi), so we multiply by itself 12 times to get it to be a real number ie: n=12.


Question 3

(iv) Arg(z) is -pi/3. We want to rotate the vector N times such that it becomes itself (perhaps with different magnitude, but we are only looking at the argument). Obviously this will occur if we go backwards by pi/3 exactly 7 times (equivalent of a 360 degree rotation). So N=7.

Question 4

(ii) The first locus is a circle centred at (0,2) with radius k. The second locus is a circle centred at (3,2) and radius 2. So the 'endpoints' of the circle occur at (1,2) and (5,2). The two circles will intersect at two points if the radius of the first locus is big enough (but not too big). Clearly this will only work if the first locus has radius *slightly* bigger than 1 or *slightly* less than 5. If you draw a sketch, you will see this clearly. I say 'slightly' because when it is equal to 1 and 5, it will intersect at only ONE point (tangential). So our answer is 1 < k < 5.

Question 5

(i)


(ii)



Question 6




Question 7

P is the same as z_1 plus z_1 rotated 90 degrees. But z_1 rotated 90 degrees is the same as z_1 times i. Therefore P = z_1 + iz_1 = (1+i)z_1


Question 8

(i) z_1 + z_2 is a vector exactly between z_1 and z_2. From the diagram, it looks like the argument will be a little bit less than 90 degrees. z_1 - z_2 is the vector pointing from z_2 to z_1.

(ii) From the question, we know that:



(iii) As I said before, the vector z_1 + z_2 is twice as long as z_1 - z_2. Draw the rhombus and label the intersection to be perpendicular (property of a rhombus). Label alpha. alpha/2 is just half the angle in between. Use SOH CAH TOA with alpha/2 to get the identity:

 
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Carrotsticks

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Carrot, for question 3, isn't it n = 7?
Yep that's right. I forgot that n=1 yields the identity vector.

I just remembered, I made this exact same mistake a couple months ago!

Somebody do Q9 for me please, I'm too tired now.

And to answer your wall-post question Accuracy, z-1 yields a real solution because when z - 1 = 0, then z = 1 (which is real).
 
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bleakarcher

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Question 1

(i) Let y=0 and solve. x=0 (double root) and x= plus/minus 2

(iii) Differentiate it using the product rule.

Question 2

(ii) z = sqrt(2) cis (pi/12). We want it to be real (this will first occur when the argument is pi), so we multiply by itself 12 times to get it to be a real number ie: n=12.


Question 3

(iv) Arg(z) is -pi/3. We want to rotate the vector N times such that it becomes itself (perhaps with different magnitude, but we are only looking at the argument). Obviously this will occur if we go backwards by pi/3 exactly 7 times (equivalent of a 360 degree rotation). So N=7.

Question 4

(ii) The first locus is a circle centred at (0,2) with radius k. The second locus is a circle centred at (3,2) and radius 2. So the 'endpoints' of the circle occur at (1,2) and (5,2). The two circles will intersect at two points if the radius of the first locus is big enough (but not too big). Clearly this will only work if the first locus has radius *slightly* bigger than 1 or *slightly* less than 5. If you draw a sketch, you will see this clearly. I say 'slightly' because when it is equal to 1 and 5, it will intersect at only ONE point (tangential). So our answer is 1 < k < 5.

Question 5

(i)


(ii)



Question 6




Question 7

P is the same as z_1 plus z_1 rotated 90 degrees. But z_1 rotated 90 degrees is the same as z_1 times i. Therefore P = z_1 + iz_1 = (1+i)z_1


Question 8

(i) z_1 + z_2 is a vector exactly between z_1 and z_2. From the diagram, it looks like the argument will be a little bit less than 90 degrees. z_1 - z_2 is the vector pointing from z_2 to z_1.

(ii) From the question, we know that:



(iii) As I said before, the vector z_1 + z_2 is twice as long as z_1 - z_2. Draw the rhombus and label the intersection to be perpendicular (property of a rhombus). Label alpha. alpha/2 is just half the angle in between. Use SOH CAH TOA with alpha/2 to get the identity:

Carrot, your not completely done with q6. You have to use geometry to find a relationship between the two arguments.
 

Carrotsticks

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Carrot, your not completely done with q6. You have to use geometry to find a relationship between the two arguments.
k = 2/3 haha.

Also next time, best not quote a huge bit of text like that. Looks a bit messy.
 

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