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Spiral similarity, for our purpose, will be a beautiful motion in the plane. It is a combination of rotation and dilation (expansion or contraction by a constant factor.

Suppose AB is a (finite) segment in the plane. CD be another segment. Assume AB and CD to be disjoint.

It is possible to *send *AB onto CD using a combination of rotation and dilation.

To accomplish this, we need to find:

- Center of rotation
- Angle of rotation
- Ratio of dilation

Join AC and BD. Suppose they intersect at X. Draw the circles CXD and AXB. Let them intersect at O.

O could be same as X or could be a different point.

**Claim: **O is the center of rotation and dilation, that sends AB to CD.

Join OA, OX, OB, OC, OD.

**Pause and think: **Can you find some cyclic quadrilaterals in this picture.

Clearly ABXO is cyclic. This implies \( \angle OBX = \angle OAX \). This is because they are subtended on the circumference by the same segment OX. (Angles subtended by a chord at the circumference are equal, by property of cyclic quadrilaterals).

Similarly CDXO is cyclic implying \( \angle ODX = \angle OCX \)

**Pause and think: **Can you find a pair of similar triangles?

*Hint: *Equiangular triangles are similar.

\( \Delta OBD \simĀ \Delta OAC \)

This is because we have shown pair of corresponding angles to be equal. Hence the two triangles are equiangular.

**Pause and imagine:**

This implies \( \angle AOC = \angle BOD \). We may imagine this as following:

- OA is turning into OC;
- OB is turning into OD

Since the angles (of rotation) \( \angle AOC, \angle BOD \) are equal, hence this works out nicely.

Moreover, OA **grows **into OC and OB **grows **into OD by same constant factor. This is true because \( \frac {OA}{OC} = \frac{OB}{OD} \). This follows from the similarity of triangles.

Hence A goes to C and B goes to D, via rotation and dilation, centered at O.

**Complex numbers** are algebraic tools to encode this motion in the plane.

(This is a supplemental note to live lecture session of Complex Number module of Cheenta I.S.I., C.M.I. Entrance Program and Math Olympiad Program)

Spiral similarity, for our purpose, will be a beautiful motion in the plane. It is a combination of rotation and dilation (expansion or contraction by a constant factor.

Suppose AB is a (finite) segment in the plane. CD be another segment. Assume AB and CD to be disjoint.

It is possible to *send *AB onto CD using a combination of rotation and dilation.

To accomplish this, we need to find:

- Center of rotation
- Angle of rotation
- Ratio of dilation

Join AC and BD. Suppose they intersect at X. Draw the circles CXD and AXB. Let them intersect at O.

O could be same as X or could be a different point.

**Claim: **O is the center of rotation and dilation, that sends AB to CD.

Join OA, OX, OB, OC, OD.

**Pause and think: **Can you find some cyclic quadrilaterals in this picture.

Clearly ABXO is cyclic. This implies \( \angle OBX = \angle OAX \). This is because they are subtended on the circumference by the same segment OX. (Angles subtended by a chord at the circumference are equal, by property of cyclic quadrilaterals).

Similarly CDXO is cyclic implying \( \angle ODX = \angle OCX \)

**Pause and think: **Can you find a pair of similar triangles?

*Hint: *Equiangular triangles are similar.

\( \Delta OBD \simĀ \Delta OAC \)

This is because we have shown pair of corresponding angles to be equal. Hence the two triangles are equiangular.

**Pause and imagine:**

This implies \( \angle AOC = \angle BOD \). We may imagine this as following:

- OA is turning into OC;
- OB is turning into OD

Since the angles (of rotation) \( \angle AOC, \angle BOD \) are equal, hence this works out nicely.

Moreover, OA **grows **into OC and OB **grows **into OD by same constant factor. This is true because \( \frac {OA}{OC} = \frac{OB}{OD} \). This follows from the similarity of triangles.

Hence A goes to C and B goes to D, via rotation and dilation, centered at O.

**Complex numbers** are algebraic tools to encode this motion in the plane.

(This is a supplemental note to live lecture session of Complex Number module of Cheenta I.S.I., C.M.I. Entrance Program and Math Olympiad Program)

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