VCE Specialist 3/4 Question Thread!

[cosine]

Hey, can some one help with this, im not 100% sure about this:


How can you factor this expression by grouping:


Also, is there any other method of solving this? Bloody hell though, these exam questions are so much different to the textbook ones.... pisses me off :/

[keltingmeith]

Hey, can some one help with this, im not 100% sure about this:


How can you factor this expression by grouping:


Also, is there any other method of solving this? Bloody hell though, these exam questions are so much different to the textbook ones.... pisses me off :/

Well - to me, it looks like there's a z - sqrt(5) + i in both of those:

[cosine]

Well - to me, it looks like there's a z - sqrt(5) + i in both of those:

omg man cant believe i didnt see that........ i need to rest, at least you wont have to answer any of my questions.
thanks by the way, #Eulereverywhere

[cosine]

Can someone show me how to solve this by the quadratic formula, or maybe complete the square?

[brightsky]

Can someone show me how to solve this by the quadratic formula, or maybe complete the square?

I think the best way to solve this equation is to use the Cartesian method.

Let z = x + yi.
(x + yi)^2 = 8 + 15i
(x^2 - y^2) + 2xyi = 8 + 15i

Equating coefficients:
x^2 - y^2 = 8...(1)
2xy = 15...(2)

Solving (1) and (2) simultaneously yields: (x,y) = (-5sqrt(2)/2, -3sqrt(2)/2) or (5sqrt(2)/2, 3sqrt(2)/2).

Hence, z = -5sqrt(2)/2 - 3sqrt(2)/2 i or z = 5sqrt(2)/2 + 3sqrt(2)/2 i.

[cosine]

Thank you so much brightsky. Is it possible however, to solve it through the polar form instead of cartesian, because theta would become a decimal?

[cosine]

Let w = 1 + ai where a is a real constant.
Show that | w^2 | = (1+ a^2 )^3/2 .


Umm..... totally blank on this question, would appreciate if someone could shed some light

[brightsky]

Thank you so much brightsky. Is it possible however, to solve it through the polar form instead of cartesian, because theta would become a decimal?

It is possible, but not advised. As you rightly mentioned, the argument is not a nice number, and so the working out might be a little messy.

Let w = 1 + ai where a is a real constant.
Show that | w^2 | = (1+ a^2 )^3/2 .


Umm..... totally blank on this question, would appreciate if someone could shed some light

I think there is something wrong with the question. The equation that you are required to prove only holds if a = 0.

w = 1 + ai
w^2 = (1+ai)^2 = 1 + 2ai - a^2 = (1-a^2) + 2ai
|w^2|
= sqrt[(1-a^2)^2 + (2a)^2]
= sqrt(1 - 2a^2 + a^4 + 4a^2)
= sqrt(1 + 2a^2 + a^4)
= sqrt[(1+a^2)^2]
= 1 + a^2 (as 1 + a^2 is always greater than 0)

The equation would hold for all a if you meant to write | w^3 | = (1+ a^2 )^3/2.

[cosine]

Thanks brightsky, i thought there was something wrong with it

= sqrt(1 + 2a^2 + a^4)
= sqrt[(1+a^2)^2]

Can you explain please how the top one became the second one? Thanks

[Stevensmay]

Thanks brightsky, i thought there was something wrong with it

= sqrt(1 + 2a^2 + a^4)
= sqrt[(1+a^2)^2]

Can you explain please how the top one became the second one? Thanks

Let

So

A nice quadratic, factorize as normal.



Substitute a back into the expression.

[cosine]

On an Argand diagram, O is the origin and P represents the point 3 + i. The point Q represents a + bi, where a and b are both positive. If triangle OPQ is equilateral, find a and b.

Please anyone!

[Zues]

just to confirm, you can only plot points that are in cartesian form, not polar, correct?

how do i transform this to cartesian and then plot

z = cis (pi/8)

[keltingmeith]

On an Argand diagram, O is the origin and P represents the point 3 + i. The point Q represents a + bi, where a and b are both positive. If triangle OPQ is equilateral, find a and b.

Please anyone!

If we have an equilateral triangle, all sides must be equal length. Since two of the sides are the mod of the two complex numbers, we have:



Of course, we need another equation - so, we consider the last length, :



Now, we just need to solve the two simultaneously:

just to confirm, you can only plot points that are in cartesian form, not polar, correct?

how do i transform this to cartesian and then plot

z = cis (pi/8)

Of course you can! Check here for an explanation on how to plot points in polar.

[Mieow]

just to confirm, you can only plot points that are in cartesian form, not polar, correct?

how do i transform this to cartesian and then plot

z = cis (pi/8)

Well actually you can. Z=cis(pi/8) tells you that the point has a magnitude of one, and is at an angle of pi/8 counterclockwise from the positive direction of the x-axis. Except when we try to plot it on the graph it's hard to figure out just from that what its coordinates are, which is why we tend to convert to cartesian form.

[cosine]

Well actually you can. Z=cis(pi/8) tells you that the point has a magnitude of one, and is at an angle of pi/8 counterclockwise from the positive direction of the x-axis. Except when we try to plot it on the graph it's hard to figure out just from that what its coordinates are, which is why we tend to convert to cartesian form.

But it would be totally okay if the value of Arg(z) was an exact value, right?