Complex Number Question (1 Viewer)

[Silences]

This question is from Patel 4U. If t is real and z = 2+it / 2-it, show that as t varies, the locus of z is a circle. Find the centre and the radius.

Thanks in advance.

 

[cyl123]

Make the denominator real:
z=(2+it)^2/[(2+it)(2-it)]=(4-t^2+4it)/(4+t^2)
Let z=x+iy

x=Re(z)=(4-t^2)/(4+t^2)
y=Im(z)=4t/(4+t^2)
The rest just involves some parametrics work in 3u to get rid of t (hint square x and y and then combine)

You should get x^2+y^2=1 i think

 

[jet]

Make the denominator real:
z=(2+it)^2/[(2+it)(2-it)]=(4-t^2+4it)/(4+t^2)
Let z=x+iy

x=Re(z)=(4-t^2)/(4+t^2)
y=Im(z)=4t/(4+t^2)
The rest just involves some parametrics work in 3u to get rid of t (hint square x and y and then combine)

You should get x^2+y^2=1 i think

Yup

 

[Silences]

Make the denominator real:
z=(2+it)^2/[(2+it)(2-it)]=(4-t^2+4it)/(4+t^2)
Let z=x+iy

x=Re(z)=(4-t^2)/(4+t^2)
y=Im(z)=4t/(4+t^2)
The rest just involves some parametrics work in 3u to get rid of t (hint square x and y and then combine)

You should get x^2+y^2=1 i think

I don't get how to use parametrics in this. I tried squaring both and it didn't work?

 

[cyl123]

I don't get how to use parametrics in this. I tried squaring both and it didn't work?

x^2=(16-8t^2+t^4)/(4+t^2)^2
y^2=16t^2/(4+t^2)^2
x^2+y^2=(16-8t^2+t^4)/(4+t^2)^2+16t^2/(4+t^2)^2
= (16+8t^2+t^4)/(4+t^2)^2
= (4+t^2)^2/(4+t^2)^2=1

 

[Silences]

x^2=(16-8t^2+t^4)/(4+t^2)^2
y^2=16t^2/(4+t^2)^2
x^2+y^2=(16-8t^2+t^4)/(4+t^2)^2+16t^2/(4+t^2)^2
= (16+8t^2+t^4)/(4+t^2)^2
= (4+t^2)^2/(4+t^2)^2=1

Oh right. Thanks.