Complex numbers locus quick question (2 Viewers)

[HeroicPandas]

 

[HeroicPandas]

Please anyone?

And its |z-(3/z)|= 5 (typed on LaTeX weirdly

 

[Carrotsticks]

Use the Reverse Triangle Inequality, then solve the quadratic for |z|.

 

[HeroicPandas]

Use the Reverse Triangle Inequality, then solve the quadratic for |z|.

I am still unsure

 

[HeroicPandas]

Can I do this after drawing an appropriate diagram

|z| + |3/z| = |z-(3/z)|

Therefore, |z| + |3/z| = 5?

 

[HeroicPandas]

Or do i split the condition and use the reverse triangle tech then solve as a quadratic?

 

[bobmcbob365]

In the diagram, you can clearly see the triangle, then using the Triangle inequality (i.e. one side is always less than the sum of the other two sides or else it wouldn't be triangle). Soz I have no idea how to use LaTeX, so i used MS Word lol.
The diagram is the 2nd one attachment.

 

[HeroicPandas]

In the diagram, you can clearly see the triangle, then using the Triangle inequality (i.e. one side is always less than the sum of the other two sides or else it wouldn't be triangle). Soz I have no idea how to use LaTeX, so i used MS Word lol.
The diagram is the 2nd one attachment.

View attachment 26631




View attachment 26630

oh ok thanks!

 

[HeroicPandas]

But can 3/z be any random complex number like z?

 

[bobmcbob365]

Well, the location of 3/z on the Argand Diagram doesn't really matter, since a triangle will be formed with z, 3/z and the origin, regardless of where, z or 3/z are.
But if you really want to know, I'm not sure if my way of thinking is correct, but if you let z = x+iy and realise 3/z, you end up with 3x/(x^2 + y^2) - i (3y/(x^2 + y^2)).
So i guess, 3/z will always be on the same side of the y-axis with z, i.e. if Re(z) is greater than zero, then Re(3/z) is also greater than zero.(z and 3/z are right of the y-axis)
Whereas, 3/z will always be on the opposite side of the x-axis with z i.e. if Im(z) is greater than zero, then Im(3/z) is less than zero. (z is over the x-axis and 3/z is under the x-axis) (because of the minus)

 

[HeroicPandas]

Well, the location of 3/z on the Argand Diagram doesn't really matter, since a triangle will be formed with z, 3/z and the origin, regardless of where, z or 3/z are.
But if you really want to know, I'm not sure if my way of thinking is correct, but if you let z = x+iy and realise 3/z, you end up with 3x/(x^2 + y^2) - i (3y/(x^2 + y^2)).
So i guess, 3/z will always be on the same side of the y-axis with z, i.e. if Re(z) is greater than zero, then Re(3/z) is also greater than zero.(z and 3/z are right of the y-axis)
Whereas, 3/z will always be on the opposite side of the x-axis with z i.e. if Im(z) is greater than zero, then Im(3/z) is less than zero. (z is over the x-axis and 3/z is under the x-axis) (because of the minus)

ahhh ok, nice thanks for the help

 

[bobmcbob365]

you're welcome =p