2011 CSSA Trial Question 2 NEED HELP (1 Viewer)

[Jennyxoxo]

CSSA 2011 Trial MX2, Question 2 C. (ii)

It asks to find the maximum value of |z|
Question 2 C, i) asks Sktech the locus of z if |z - 3 + 3i| </= 2
I got the Q2, C, i but dont get c, ii


Need your help,

Thanks everyone

 

[Johnstan]

the max value is OC + CD from ur locus diagram ( where c is the centre, and D is a point on the external.. radius is 2)
so OC=3ROOT2 AND CD=2 ( radius)
so max l z l = 3root2 +2

 

[Jennyxoxo]

yeah but is that like a standard/ fixed rule???

Like the max always goes through the origin and centre???

 

[Johnstan]

for max modulus yeah- its the addition of the length from origin to the centre and the radius

 

[Jennyxoxo]

for max modulus yeah- its the addition of the length from origin to the centre and the radius

Thanks <3

 

[Johnstan]

Thanks <3

no worries
u doing the exam tomorrow?

 

[Carrotsticks]

If you don't have a diagram, you can still find the maximum/minimum Modulus using the Triangle Inequality or the Reverse Triangle Inequality.

 

[Johnstan]

If you don't have a diagram, you can still find the maximum/minimum Modulus using the Triangle Inequality or the Reverse Triangle Inequality.

yeahps, triangle inequality is bauss.
but in this particular question they asked for a locus on argand in part i) so much easier to get it that way imho.