Complex Help (1 Viewer)

Intilegience · Dec 18, 2023

Can you guys help me with how to do this question?

17. the points P1, P2 and P3 represent the complex numbers z1, z2 and z3 respectively. If z2/z1 = z3/z2 , show that OP2 bisects angle P1OP3.

scaryshark09 · Dec 18, 2023

Intilegience · Dec 18, 2023

Thank you so much!

Luukas.2 · Dec 18, 2023

Or, a longer way...

$\begin{align*} \frac{z_2}{z_1} &= \frac{z_3}{z_2} \\ \arg{\left(\frac{z_2}{z_1}\right)} &= \arg{\left(\frac{z_3}{z_2}\right)} \\ \arg{z_2} - \arg{z_1} &= \arg{z_3} - \arg{z_2} \\ \arg{z_2} &= \frac{1}{2}\left(\arg{z_1} + \arg{z_3}\right) \\ \\ \text{Thus,} \quad \angle{P_2OX} &= \frac{1}{2}\left(\angle{P_1OX} + \angle{P_3OX}\right) = \frac{\alpha + \beta}{2} \qquad \text{where $\alpha = \angle{P_1OX}$ and $\beta = \angle{P_3OX}$ and $X$ is on the positive real axis} \\ \\ \text{Now,} \quad \angle{P_3OP_2} &= \angle{P_3OX} - \angle{P_2OX} = \beta - \frac{\alpha + \beta}{2} = \frac{\beta - \alpha}{2} \\ \text{And,} \quad \angle{P_2OP_1} &= \angle{P_2OX} - \angle{P_1OX} = \frac{\alpha + \beta}{2} - \alpha = \frac{\beta - \alpha}{2} \\ \\ \text{Thus,} \quad \angle{P_3OP_2} &= \angle{P_2OP_1} \qquad \text{and $OP_2$ bisects $\angle{P_3OP_1}$ as required} \end{align*}$

Susu123 · Dec 18, 2023

Intilegience said:
Can you guys help me with how to do this question?

17. the points P1, P2 and P3 represent the complex numbers z1, z2 and z3 respectively. If z2/z1 = z3/z2 , show that OP2 bisects angle P1OP3.

is this from cambridge

Complex Help (1 Viewer)

Intilegience

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