VCE Specialist 3/4 Question Thread!

[BubbleWrapMan]

Similarly, if you write out you can "by inspection" work out the other factor, by imagining what the expanded product of the two factors would look like. (Exact same method as that of Climbtoohigh, but could be used if you can manage the numbers in your head/scrap piece of paper, could possibly be quicker.)

It's somewhat risky to do that on an exam 1, though in retrospect I guess it was somewhat easy to see that b = 3.

[Homer]

Assuming , then , so the result holds iff

thanks for your reply russ, but im still a little confused :/

[BubbleWrapMan]

Was it meant to be rather than ? It's a question from a VCAA exam.

[Hancock]

ClimbTooHigh is right. I remember the question.

[jadams]

thanks for your reply russ, but im still a little confused :/

Russ is saying that the statement is only true/proven if only holds true when a=0, which would answer the question you posted.

Most likely though, the question is a little different than what you posted.

[Homer]

I think there is a printing mistake as climbtoohigh and hancock said its ment to be w^3 instead of w^2, but thanks for your help guys

[Homer]

just a theoretical question related to relations and regions of the complex plane. When drawing rays suppose {z:arg(z)=theta} why is there an open circle at the origin, showing 0+0i as not a member of the set?

[BubbleWrapMan]

You can't define the argument of zero, it's like trying to define the direction of the zero vector.

[Homer]

if the ray was translated suppose {z:arg(z+a+bi)=theta,a,b belong to R} would the point (-a-bi) be a member of the set?

[BubbleWrapMan]

If you translate an undefined point it's still undefined

[Homer]

yeh, coz i got confused since the book didnt have an open circle :/

[BubbleWrapMan]

Is wrong

[Homer]

when converting {z: |z+3|+ |z+1|=4}  into Cartesian form, i keep getting , but I can never transform it to  as shown in the answer :/

[brightsky]

complete the square

[BubbleWrapMan]

(squared both sides)