Question

Let z1,z2,and z3 be complex numbers such that |z1|=|z2|=|z3|=r>0 and z1+z2+z3≠0 . Prove that |z1z2+z2z3+z3z1÷z1+z2+z3| = r

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=r>0\,\,\,\,\,and}\\\tt{z_{1}+z_{2}+z_{3}\ne0}$

$\bf{Let\,\,z_{1}=r\,e^{i\theta_{1}},\,\,z_{2}=r\,e^{i\theta_{2}},\,\,z_{3}=r\,e^{i\theta_{3}}}$

Now,

$\sf{\left|\dfrac{z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}}{z_{1}+z_{2}+z_{3}}\right|}$

$\sf{=\left|\dfrac{r\,e^{i\theta_{1}}\cdot\,r\,e^{i\theta_{2}}+r\,e^{i\theta_{2}}\cdot\,r\,e^{i\theta_{3}}+r\,e^{i\theta_{3}}\cdot\,r\,e^{i\theta_{1}}}{r\,e^{i\theta_{1}}+r\,e^{i\theta_{2}}+r\,e^{i\theta_{3}}\right|}$

$\sf{=\left|\dfrac{r^2\cdot\,e^{i\theta_{1}}\cdot\,e^{i\theta_{2}}+r^2\cdot\,e^{i\theta_{2}}\cdot\,e^{i\theta_{3}}+r^2\cdot\,e^{i\theta_{3}}\cdot\,e^{i\theta_{1}}}{r\left(e^{i\theta_{1}}+e^{i\theta_{2}}+e^{i\theta_{3}\right)}\right|}$

$\star\bf{\,We\,\,know,\,\,for\,\,any\,\,complex\,\,number\,\,z,\,\,|z|^2=z\cdot\,\bar{z}}$

$\mapsto\bf{\,\,Since\,\,|z_{1}|=|z_{2}|=|z_{3}|=r}\\\\\mapsto\bf{\,\,r^2=z_{1}\cdot\bar{z}_{1},\,\,r^2=z_{2}\cdot\bar{z}_{2},\,\,r^2=z_{3}\cdot\bar{z}_{3}}$

So,