# Locus q (1 Viewer)

#### =)(=

##### Active Member

I don't quite fully understand part b and c

#### jimmysmith560

##### Le Phénix Trilingue
Moderator
Would the following working help?

$\bg_white w=1+3i$

$\bg_white \therefore \left|w\right|=\sqrt{10}$

Part (a):

Since $\bg_white \left|z\right|=\left|w\right|$

$\bg_white \therefore \left|z\right|=\sqrt{10}$

This is a circle with centre the origin and the radius equal to the modulus of w, as follows:

Part (b):

This is a circle with centre w and the radius equal to twice the modulus of w, as follows:

Part (c):

This is the perpendicular bisector of the line segment through w and the origin, as follows:

#### =)(=

##### Active Member
Would the following working help?

$\bg_white w=1+3i$

$\bg_white \therefore \left|w\right|=\sqrt{10}$

Part (a):

Since $\bg_white \left|z\right|=\left|w\right|$

$\bg_white \therefore \left|z\right|=\sqrt{10}$

This is a circle with centre the origin and the radius equal to the modulus of w, as follows:

View attachment 35507

Part (b):

This is a circle with centre w and the radius equal to twice the modulus of w, as follows:

View attachment 35508

Part (c):

This is the perpendicular bisector of the line segment through w and the origin, as follows:

View attachment 35509

#### yanujw

##### Active Member
Interpret this as |z-w| = |z-0|, which means the distance from w must be the same as the distance from the point 0. The midpoint of 0 and w (0.5 + 1.5i) is one such point, and it follows that the line perpendicular line passing through this point retains this property.

#### vishnay

##### God
just substitute in $\bg_white z=x+iy$ if ur ever unsure