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Let z be a complex number such that $\left | \frac{z-i}{z+2i} \right |=1$ and $\left | z \right |=\frac{5}{2}$ . Then the value of $\left | z+3i \right |$ is :

Option: 1
$\sqrt{10}$

Option: 2
$\frac{7}{2}$

Option: 3
$\frac{15}{4}$

Option: 4
$2\sqrt{3}$

Answers (1)

best_answer

Distance formula, Equation of perpendicular bisector -

Distance between two points A(z₁) and B(z₂) can be found as

AB = |z₂ - z₁| = | Affix of B - Affix of A |

$\\\mathrm{Let \; z_1 = x + iy \;and\; z_2 = x + iy } \\\mathrm{Then\;,\left | z_1 - z_2 \right | = \left | (x_1 - x_2) + i(y_1-y_2) \right |} \\\mathrm{=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}} \\\mathrm{= distance \;between \;points\; (x_1, y_1) \; and} \\\mathrm{ (x_2, y_2) = distance \;between\; between \;z_1 \; and \; z_2} \\\mathrm{where\; z_1 = x_1 + iy_1\; and \; z_2 = x_2 + iy_2 }$

The distance of a point from the origin is |z - 0| = |z|
Three points A(z₁), B(z₂) and C(z₃) are collinear, then AB + BC = AC

i.e. |z₂ - z₁| + |z₃- z₂| = |z₃ - z₁|

We can use the distance formula to find the equation of perpendicular bisector formula.

Let two fixed points A(z₁) and B(z₂) and a moving point C(z) which is always at the same

Distance from A(z₁) and B(z₂)

AC = BC

Locus of a point equidistant from two given points.

|z - z₁| = |z - z₂|

z will lie on the perpendicular bisector of the line joining $z_{1}$ and $z_{2}$ .

This is the equation of perpendicular of bisector as z always remains at the same distance from

A(z₁) and B(z₂)

-

Area of triangle, circle (formula) -

Equation of Circle:

The equation of the circle whose center is at the point $z_0$ and have radius r is given by

$|z-z_0| = r$

If the center is origin then, $z_0=0$ , hence equation reduces to |z| = r

Interior of the circle is represented by $|z-z_0| < r$

The exterior is represented by $|z-z_0| > r$

Here z can be represented as x + iy and $z_0$ is represented by $x_0 + iy_0$

-

$\\|z-i|=|z+2i|\\this \;is\;the\;equation\;of\;perpendicular\;bisector$

z = $(\sqrt6,-1,2)$

$\\z=\sqrt 6-1/2i\\|z+3i|=|\sqrt6+(3-1/2)i|=\sqrt{6+\frac{25}{4}}=\frac{7}{2}\\$

Posted by

Devendra Khairwa

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