alternative method? (1 Viewer)

[rickypontingdagoat]

hey I was wondering if you could do part ii) using triangle inequalities instead of graphically? I know the first part is doable.

Thanks.

 

[WeiWeiMan]

|z-3i+(-2+2i)|=1
1|≤|z-3i|+|-2+2i|
i could be completely wrong but that should give a rough idea
how are you 2028

 

[rickypontingdagoat]

I'm not, how are you 2026?

 

[WeiWeiMan]

I'm not, how are you 2026?

cuz i'm literally 2026

 

[Average Boreduser]

Just realised u meant p2 mb

 

[Average Boreduser]

still don't thing triangle law will be that helpful. if u found the min max for |z|, you can look at z-3i, makes a triangle, my guess it might be isosceles then the max might be a similar triangle? using ratio find min max? I'm blurting out crap rn bc im scared im gonna get it wrong.

 

[Drongoski]

???

 

[Luukas.2]

View attachment 41999
hey I was wondering if you could do part ii) using triangle inequalities instead of graphically? I know the first part is doable.

Thanks.

Yes, you can.

For part (ii), take and .

We are given that and can calculate that .

We seek .

Applying the triangle inequality gives us that:

​

And similarly, that:

​

Combining these:

​

 

[Luukas.2]

Yes, you can.

For part (ii), take and .

We are given that and can calculate that .

We seek .

Applying the triangle inequality gives us that:

​

And similarly, that:

​

Combining these:

​

Note, also, that the method generalises, to:

​

is the result without any substitutions.

Applying it to part (i), we use:

​

and so, by substitution, we can confirm @Drongoski's answer, that

​

We can also expand the method to a circle with any centre and any other point on the Argand Diagram, . This can provide a quick check for analogous MX1 vector problems.

It can even be applied to cases like , which actually lies on the locus , as:

​

and so yields the (expected) result, that the distance must be between zero and the length of the diameter:

​

 

[rickypontingdagoat]

Thank you so much!