complex number loci

[ruby osprey]

how would i do part ci and cii?

[undone glade]

for ci, find both roots of the equation

sub each root into the inequality and check to see if it's true

[ruby osprey]

yh 1/2 + 3/2 i

and its conjugate

[undone glade]

yeah

put those into that inequality

|z-2|<sqrt5

[ruby osprey]

for z?

[undone glade]

yeah

[ruby osprey]

ohh

but how would ik z represents the quadratic

because it never mentioned it anywhere

i thought z is just a complex number

[undone glade]

the quadratic is written in terms of z

2z^2-2z+5

z is like any other letter

x, y, etc

[ruby osprey]

yhh

just thought x would represent a +bi

but if its the quadratic

fair enough

and what would i do for cii

[undone glade]

|z|<|z-2i| represents all points below the line 2i

[ruby osprey]

oh yh

mb

[undone glade]

so the root satisfying the inequality must have an imaginary part less than 2i

[ruby osprey]

i was thinking it was the point on the line for some reaosn

yh my bad

thank you tho

[undone glade]

np

[glad vortex]

for cii

nvm

[glad vortex]

btw

in a case where say none of the roots would satisfy

youd also check the distance between the centre and the root right

and p rove its greater than the radius

[undone glade]

yeah

[undone glade]

it's easiest to draw everything on an argand diagram tbh

[glad vortex]

yh