MHB Locus of Line Segment Mid Point Intercept Real/Imaginary Axis

[juantheron]

Find locus of mid point of line segment intercept between real and imaginary axis by the line

$a\bar{z}+\bar{a}z+b=0,$ where $b$ is areal parameter and $a$ is a fixed complex

number such that $\Re(a),\Im(a)\neq 0$

My Attempt:: Let $z=x+iy$ and $a=x_{0}+iy_{0}$, Then put into $a\bar{z}+\bar{a}z+b=0,$

We get $(x_{0}+iy_{0})(x-iy)+(x_{0}-iy_{0})(x+iy)+b=0$

So $(2xx_{0}+2yy_{0}+b)+i(2xy_{0}+2x_{0}y) = 0+i\cdot 0$

So $2xx_{0}+2yy_{0}+b=0$ and $2xy_{0}+2x_{0}y=0$

Now How can i solve it after that, Help me, Thanks

 

[Evgeny.Makarov]

Re: Locus of line sagment

We get $(x_{0}+iy_{0})(x-iy)+(x_{0}-iy_{0})(x+iy)+b=0$

So $(2xx_{0}+2yy_{0}+b)+i(2xy_{0}+2x_{0}y) = 0+i\cdot 0$

You have an error in signs. The imaginary part should be 0 because $a\bar{z}+\bar{a}z=a\bar{z}+\overline{a\bar{z}}=2\text{Re}(a\bar{z})$.

 

[HallsofIvy]

Re: Locus of line sagment

Find locus of mid point of line segment intercept between real and imaginary axis by the line

$a\bar{z}+\bar{a}z+b=0,$ where $b$ is areal parameter and $a$ is a fixed complex

number such that $\Re(a),\Im(a)\neq 0$

My Attempt:: Let $z=x+iy$ and $a=x_{0}+iy_{0}$, Then put into $a\bar{z}+\bar{a}z+b=0,$

We get $(x_{0}+iy_{0})(x-iy)+(x_{0}-iy_{0})(x+iy)+b=0$i

$(x_0x0- ix_0y+ ixy_0+ y_0y)+ x_0x+ ix_0y- iy_0x+ y_0y)+ b= 0$
$(2x_0x+ 2y_0y)+ i(-x_0y+ xy_0+ x_0y- y_0x)= 2x_0x+ 2y_0y= 0$

So $(2xx_{0}+2yy_{0}+b)+i(2xy_{0}+2x_{0}y) = 0+i\cdot 0$

So $2xx_{0}+2yy_{0}+b=0$ and $2xy_{0}+2x_{0}y=0$

Now How can i solve it after that, Help me, Thanks

 

[soroban]

Find locus of midpoint of line segment intercept between real and imaginary axis by the line $a\bar{z}+\bar{a}z+b=0,$
where $b$ is a real parameter and $a$ is a fixed complex number such that $\Re(a),\Im(a)\neq 0$

Let $z=x+iy$ and $a=x_o+iy_o$.

Substitute into $a\bar{z}+\bar{a}z+b=0:$
. . $(x_o+iy_o)(x-iy)+(x_o-iy_o)(x+iy)+b\:=\:0$

Expand:
. . . $x_ox - ix_oy + ixy_o + y_oy + x_ox + ix_oy - ixy_o + y_oy + b \:=\:0$

We have: $2x_ox + 2y_oy + b \:=\:0 $

The intercepts are $\left(-\dfrac{b}{2x_o}, 0\right),\;\left(0, -\dfrac{b}{2y_o}\right)$

Their midpoint is: $\left(-\dfrac{b}{4x_o},\,-\dfrac{b}{4y_o}\right)$