Given z1,z2,z3z_1,z_2,z_3 and z4z_4 are complex numbers, prove that the line joining z1,z2z_1,z_2 and the line joining from z3,z4z_3,z_4 are perpendicular iff Re{(z1−z2)(ˉz3−ˉz4)}=0Re\{(z_1-z_2)(\bar z_3-\bar z_4)\}=0. Try not to use polar form.

I try to start with writing Re{(z1−z2)(ˉz3−ˉz4)}=Re{z1ˉz3}−Re{z1ˉz4}−Re{z2ˉz3}+Re{z2ˉz4}Re\{(z_1-z_2)(\bar z_3-\bar z_4)\}=Re\{z_1\bar z_3\}-Re\{z_1\bar z_4\}-Re\{z_2\bar z_3\}+Re\{z_2\bar z_4\} (I’m not sure if it’s right)

Then how can I make use of the perpendicular condition? Any hints for the reverse direction, or I just have to reverse the argument?

Thank you!

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If vv and ww are orthogonal vectors then v⋅w=0v\cdot w=0. This is the definition of orthoganility.

– Masacroso

2 days ago

how do we define dot product on a complex plane? And why do we have to take conjugate of z3z_3 and z4z_4?

– mathshungry

2 days ago

1

The euclidean dot product on Cn\Bbb C^n is defined as v⋅w:=n∑k=1vkˉwkv\cdot w:=\sum_{k=1}^n v_k \bar{w}_k See the definition and characteristics of any dot product here.

– Masacroso

2 days ago

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1 Answer

1

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Let z1−z2=aeiθz_1-z_2=ae^{i\theta} and z3−z4=beiϕz_3-z_4=be^{i\phi}, then

(z1−z2)¯(z3−z4)=abei(θ−ϕ)(z_1-z_2)\overline{(z_3-z_4)} = abe^{i(\theta-\phi)}

Re[(z1−z2)¯(z3−z4)]=abcos(θ−ϕ)\operatorname{Re}[(z_1-z_2)\overline{(z_3-z_4)}] = ab\cos (\theta-\phi)

Re[(z1−z2)¯(z3−z4)]=0⟺cos(θ−ϕ)=0⟺θ−ϕ=(n+12)π⟺(z1−z2)⊥(z3−z4)\begin{align*}

\operatorname{Re}[(z_1-z_2)\overline{(z_3-z_4)}]=0

& \iff \cos (\theta-\phi) =0 \\

& \iff \theta-\phi=\left(n+\frac{1}{2} \right) \pi \\

& \iff (z_1-z_2)\perp (z_3-z_4)

\end{align*}

where aa, b>0b> 0