VCE Specialist 3/4 Question Thread!

[psyxwar]

Think you should revisit your index and log laws lol

i^40 = (i^4)^10 = 1

[psyxwar]

It's the same thing really. In your previous post you had i^4*i^10=i^40, which is false. It equals i^14.

For these questions just split it up into multiples of four and cancel that bit out as it equals one.

For example with i^9 you get (i^4)^2*i=i

[Zues]

how do i do the "b" part?

[brightsky]

zw = (a+bi)(c+di) = ac + adi + bci - bd = (ac - bd) + (ad + bc) i
Hence, LHS = (ac - bd) - (ad + bc) i.
Now, RHS = (a - bi)(c - di) = ac - adi - bci - bd = (ac - bd) - (ad + bc) i.
Hence, LHS = RHS, as required.

[Zues]

yeap i see, thanks!

i thougth it would be something else , didnt process z = .... and w = ... correctly!

[keltingmeith]

anyone??

Pretty sure that should be positive, I can't see how it could be negative.

[brightsky]

Note in part c), 1/2(z + conj(z)) = Re(z) and 1/2(z-conj(z)) = Im(z) i. In part b), z = 3 + 2sqrt(2) i. So 1/2(z - conj(z)) = Im(z) i = 2sqrt(2) i. It follows from this that Heinemann is a dodgy textbook.

As an aside, the identities 1/2(z + conj(z)) = Re(z) and 1/2(z-conj(z)) = Im(z) i are critical and should be memorised. At some point in your mathematical career, you will learn about Euler's identity: e^(ix) = cos(x) + isin(x). Note that if we let z = e^(ix) = cos(x) + isin(x), then Re(z) = cos(x) = 1/2(z + conj(z)) = 1/2(e^(ix) + e^(-ix)) and Im(z) = sin(x) = 1/(2i)(z - conj(z)) = 1/(2i)(e^(ix) - e^(-ix)). The identities cos(x) = 1/2(e^(ix) + e^(-ix)) and sin(x) = 1/(2i)(e^(ix) - e^(-ix)) you will find to be incredibly useful in a range of applications.

[ikiwi]

Arg(z) is the same as arg(z) except the angle must be restricted to (-pi,pi].
For the picture, they are asking for arg(z) - note the lower case a - so the angle does not need to be restricted. Also, the argument is taken from the positive direction of the Re(z) axis (the right side).

[ikiwi]

See the attached picture.
The angle for the argument you want is red - from the positive x direction.
The angle pi/4 is in the first quadrant, to get the argument, you need to add pi (half a circle) to get the angle in the 3rd quadrant. In order to do so, you add the blue angle to the green angle.

[Zues]

See the attached picture.
The angle for the argument you want is red - from the positive x direction.
The angle pi/4 is in the first quadrant, to get the argument, you need to add pi (half a circle) to get the angle in the 3rd quadrant. In order to do so, you add the blue angle to the green angle.

why is it necessary to have it in the third quadrant.

Plus can someone explain this
For non-zero z an infinite number of arguments of z exist since, for a
given z, any of the angles θ ± 2nπ, n ∈ N also represents the position of
point P in the figure above because a clockwise or anticlockwise rotation
consisting of multiples of 2π radians (or 360°) merely moves P to its original position.
To ensure that there is only one value of θ corresponding to z we refer to the principal value of θ.
The principal value is the angle θ in the range −π < θ ≤ π or θ ∈ (−π, π].

in relation to the screenshot would be awesome!

they say it has to be between -pi and pi thus isnt this the whole circle? one half loop anti clockwise and one clockwise?
and wouldnt adding / subtracting 2pi get you in the same position?

[Zues]

and i thought lowercase a for arg was not restricted?

[ikiwi]

why is it necessary to have it in the third quadrant.

Because you're finding the argument of which, if you draw it on the argand diagram is in the 3rd quadrant.

they say it has to be between -pi and pi thus isnt this the whole circle? one half loop anti clockwise and one clockwise?
and wouldnt adding / subtracting 2pi get you in the same position?

Yes, the argument has to encompass the whole circle because you can have complex numbers with any angle. Yes, adding/subtracting 2pi gets you to the same position, which is the whole point of doing so. Its just that, since we're using a circle, its easier to deal with the simplest form of the angle as possible (kind of like simplifying fractions). The simplest form of the angle is between -pi and pi, which is why the principal argument has this domain restriction. We're not changing the complex number or angle in anyway, we are just trying to simplify the number to make it easier for more complex manipulations later.

and i thought lowercase a for arg was not restricted?

Yes, arg is not restricted but Arg is. arg is just called the argument and Arg is the principal argument.

[Zues]

Because you're finding the argument of which, if you draw it on the argand diagram is in the 3rd quadrant.
Yes, the argument has to encompass the whole circle because you can have complex numbers with any angle. Yes, adding/subtracting 2pi gets you to the same position, which is the whole point of doing so. Its just that, since we're using a circle, its easier to deal with the simplest form of the angle as possible (kind of like simplifying fractions). The simplest form of the angle is between -pi and pi, which is why the principal argument has this domain restriction. We're not changing the complex number or angle in anyway, we are just trying to simplify the number to make it easier for more complex manipulations later.
Yes, arg is not restricted but Arg is. arg is just called the argument and Arg is the principal argument.

ok i get the first bit
second bit. exactly why do we want to subtract or add 2 pi if it gets us in the same position. How does doing so give us the simplest angle possible? If theta has to be greater then -pi but also equal to and less then pi then this covers the whole circle, when do we know when to add or subtract 2 pi?

[keltingmeith]

ok i get the first bit
second bit. exactly why do we want to subtract or add 2 pi if it gets us in the same position. How does doing so give us the simplest angle possible? If theta has to be greater then -pi but also equal to and less then pi then this covers the whole circle, when do we know when to add or subtract 2 pi?

... I'm honestly confused by your question.

You add or subtract 2pi so that you can get your Argument into the desired domain of (-pi, pi]. Obviously if your angle is lower than that domain, then you'd add 2pi, and if it's above that domain, you'd subtract 2pi.

[Zues]

... I'm honestly confused by your question.

You add or subtract 2pi so that you can get your Argument into the desired domain of (-pi, pi]. Obviously if your angle is lower than that domain, then you'd add 2pi, and if it's above that domain, you'd subtract 2pi.

you answered it, cheers

Ive got two questions

can someone show me how to "locate z = 5cis ( - pi / 4) and express in Cartesian form"

also in the attached, why does it become negative in the -1/squrt2