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avatar+2578 

Can anyone help me with this one?

 Nov 24, 2021
 #1
avatar+117447 
+3

Hi Solveit  laugh

 

I let   z=x+yi

and this is the graph I got.

 

I don;t intend to do all the working but I will say that

 

\(|z+4+3i|\\ =|(x+4)+(y+3)i|\\ =(x+4)^2+(y+3)^2\\ etc\)

 

see if you can do it from there,     laugh

 

 Nov 25, 2021
 #2
avatar+2578 
+3

So I came till here, but don't know how to continue:

\(|x+yi+4+3i|\leq|5\cdot(x+yi)+10-9i|\\ |(x+4)+i\cdot(y+3)|\leq|(5x+10)+i\cdot(5y-9)|\\ \sqrt{(x+4)^2+(y+3)^2}\leq\sqrt{(5x+10)^2+(5y-9)^2}\\ x^2+8x+16+y^2+6y+9\leq25x^2+100x+100+25y^2-90y+81\\ x^2+8x+y^2+6y+25\leq25x^2+100x+25y^2-90y+181\\ -24x^2-24y^2-92x+96y\leq156\)

Solveit  Nov 25, 2021
 #3
avatar+117447 
+2

Nice work Solviet.

 

Divide everything by  -24

then complete the squares to get    (x-h)^2 + (y-k)^2 >= r^2

 Nov 25, 2021

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