# Complex no.

• Feb 13th 2011, 03:06 AM
jacks
Complex no.
If $\displaystyle Z_{1},Z_{2}$ and $\displaystyle Z_{3}$ are Complex no. Representing The Vertices of a Triangle inscribed in $\displaystyle |Z|=2$. The

Altitude Through $\displaystyle Z_{1}$ meets the Circumcircle in $\displaystyle P$, Then The Complex no. Corrosponding To $\displaystyle P$ is =

(Ans in Terms of $\displaystyle Z_{1},Z_{2}$ and $\displaystyle Z_{3}$)
• Feb 13th 2011, 03:58 PM
Sudharaka
Quote:

Originally Posted by jacks
If $\displaystyle Z_{1},Z_{2}$ and $\displaystyle Z_{3}$ are Complex no. Representing The Vertices of a Triangle inscribed in $\displaystyle |Z|=2$. The

Altitude Through $\displaystyle Z_{1}$ meets the Circumcircle in $\displaystyle P$, Then The Complex no. Corrosponding To $\displaystyle P$ is =

(Ans in Terms of $\displaystyle Z_{1},Z_{2}$ and $\displaystyle Z_{3}$)

Dear jacks,

Let, $\displaystyle z_1=x_1+iy_1$

$\displaystyle z_2=x_2+iy_2$

$\displaystyle z_3=x_3+iy_3$

Then write the gradient of the $\displaystyle Z_{2}Z_{3}$ line. Let it be $\displaystyle m_1$.

Since $\displaystyle Z_{2}Z_{3}$ and $\displaystyle PZ_{1}$ are perpendicular you could find the gradient of the $\displaystyle PZ_{1}$ line by,

$\displaystyle m_{1}\times m=-1$

Then you could fine the equation of the $\displaystyle PZ_{1}$ line. Hint: You know the gradient and a point on the $\displaystyle PZ_{1}$ line. Let the equation be $\displaystyle y=mx+c$----(1)

The equation of the circle could be written as, $\displaystyle \mid Z\mid=2\Rightarrow{x^2+y^2=4}$-----------(2)

Using equations (1) and (2) you can find the point P.