# Short cuts in COMPLEX NUMBERS

Here goes the revision tips for a quick recollection of the important content in the chapter COMPLEX NUMBERS.

Equation of perpendicular bisector: The equation of the perpendicular bisector of the line segment joining points z₁ and z₂ isz (z₁̅ – z₂̅) + z̅ (z₁ – z₂) = |z₁|² – |z₂|².

Important results: If z₁, z₂, z₃ are the points A, B and C in argand plane then

$$\angle BAC=\arg \left( \frac{{{z}_{3}}-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}} \right)$$.

$$\frac{{{z}_{3}}-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}=\left| \frac{{{z}_{3}}-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}} \right|(\cos \alpha +i\sin \alpha ),\alpha =\angle BAC$$

Two triangles z₁, z₂, z₃ are vertices of ABC and a₁, a₂, a₃ vertices of DEF are similar if  $$\left| \begin{matrix}{{z}_{1}} & {{a}_{1}} & 1\\{{z}_{2}} & {{a}_{2}} & 1 \\{{z}_{3}} & {{a}_{3}} & 1 \\\end{matrix} \right|=0$$.

The equation of circle having z₁ and z₂ as the end points of the diameter is (z – z₁) (z̅ – z̅₂) + (z̅ – z̅₁) (z – z₂)

If z₁ and z₂ are fixed complex numbers then locus of a point Z satisfying $$\arg \left( \frac{z-{{z}_{1}}}{z-{{z}_{2}}} \right)=\pm \frac{\pi }{2}$$ is a circle having z₁, z₂ as end points of diameter.(i) The equation of circle passing through 3 points z₁, z₂, z₃ is $$\left( \frac{z-{{z}_{3}}}{z-{{z}_{1}}} \right)\left( \frac{{{z}_{2}}-{{z}_{1}}}{{{z}_{2}}-{{z}_{3}}} \right)=\left( \frac{\overline{z}-\overline{{{z}_{3}}}}{\overline{z}-\overline{{{z}_{1}}}} \right)\left( \frac{\overline{{{z}_{2}}}-\overline{{{z}_{1}}}}{\overline{{{z}_{2}}}-\overline{{{z}_{3}}}} \right)$$.

(ii) If four points z₁, z₂, z₃ z₄ are concyclic $$\left( \frac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}-{{z}_{4}}} \right)\left( \frac{{{z}_{3}}-{{z}_{4}}}{{{z}_{3}}-{{z}_{2}}} \right)$$ must be purely real.

Some standard loci in the argand plane: If z is a variable point and z₁, z₂ are two fixed points in the argand plane then

(i) |z – z₁| = |z – z₂|

Locus of z is perpendicular bisector of line segment joining z₁ and z₂.

(ii) |z – z₁| + |z – z₂|=|z₁ – z₂|

Locus of z is line segment joining z₁ and z₂.

(iii) |z – z₁| + |z – z₂|= k (≠|z₁ – z₂|) locus of z is ellipse.

(iv) |z – z₁| – |z – z₂|= k |z₁ – z₂|.

Locus of z is a straight line joining z₁ and z₂ but does not lie between z₁ and z₂.

(v) |z – z₁| – |z – z₂|= k (≠|z₁ – z₂|).

Locus of z is hyperbola.

(vi) |z – z₁|² + |z – z₂|²= |z₁ – z₂|².

Locus of z is a circle with z₁ and z₂ as extremities of diameter.

(vii) |z – z₁| = k |z – z₂|   k ≠ 1.

Locus of z is a circle.

(viii) $$\arg \left( \frac{z-{{z}_{1}}}{z-{{z}_{2}}} \right)=\alpha \,(fixed)$$.
Locus of z is segment of circle

(ix) $$\arg \left( \frac{z-{{z}_{1}}}{z-{{z}_{2}}} \right)=0\,(or)\pi$$.

Then locus of z is straight line through z₁, z₂.

If z₁, z₂, z₃ are vertices of a triangle ABC then its orthocenter is $$=\frac{(a\sec A){{z}_{1}}+(b\sec B){{z}_{2}}+(c\sec C){{z}_{3}}}{a\sec A+b\sec B+c\sec C}$$.

The circumcenter of with z₁, z₂, z₃ as vertices of triangle is $$z=\frac{\sum{{{z}_{1}}\overline{{{z}_{1}}}({{z}_{2}}-{{z}_{3}})}}{\sum{\overline{{{z}_{1}}}({{z}_{2}}-{{z}_{3}})}}$$.

Orthocenter will be $$z=\frac{\sum{{{z}^{2}}_{1}(\overline{{{z}_{2}}}-\overline{{{z}_{3}}})}+\sum{{{z}_{1}}^{2}({{z}_{2}}-{{z}_{3}})}}{\sum{({{z}_{1}}\overline{{{z}_{2}}}-\overline{{{z}_{1}}}{{z}_{2}})}}$$.

If z₁, z₂, z₃ are vertices of a triangle, if the triangle is equilateral then $$\frac{1}{{{z}_{2}}-{{z}_{3}}}+\frac{1}{{{z}_{3}}-{{z}_{1}}}+\frac{1}{{{z}_{1}}-{{z}_{2}}}=0$$.

$${{z}^{2}}_{1}+{{z}_{2}}^{2}+{{z}_{3}}^{3}={{z}_{1}}{{z}_{2}}+{{z}_{2}}{{z}_{3}}+{{z}_{3}}{{z}_{1}}$$.

Let zₒ be circumcentre then $${{z}^{2}}_{1}+{{z}_{2}}^{2}+{{z}_{3}}^{3}=3{{z}^{2}}_{0}$$.