Let A1 and A2 represent z1 and z2. Statement-1: Equation of circle with A1 A2 as diameter is |2

Let A1 and A2 represent z1 and z2. Statement-1: Equation of...

Key Concepts

Orthogonality in Complex Numbers

Orthogonality, or the concept of perpendicularity between vectors, has an algebraic counterpart in the complex plane. It frequently shows up in conditions where the dot product (or its equivalent using complex numbers and their conjugates) is zero. This concept is essential when deducing geometric properties of figures constructed using complex numbers, such as verifying that a given point on a circle forms a right angle with the endpoints of a diameter.

Thale's Theorem and Right Angle Inscription

Thale's theorem states that any angle inscribed in a semicircle is a right angle. When working with circles defined by a diameter from two complex numbers, this theorem underpins the result that any point on the circle (other than the endpoints of the diameter) forms a right angle with the diameter. This property is often encoded algebraically using relationships involving complex numbers and their conjugates.

Circle in the Complex Plane

This concept involves representing geometric figures such as circles using complex numbers. In particular, an equation of a circle can be written in terms of the modulus of a complex expression, indicating all points (represented as complex numbers) that are at a fixed distance (the radius) from a given center. This method provides an elegant link between algebra and geometry.

Complex Conjugation

Complex conjugation is an operation on complex numbers that reflects a point across the real axis. It is a crucial tool in complex geometry as it allows one to express and manipulate distances and angles. Particularly, the product of a complex number and the conjugate of another appears frequently in deriving geometric conditions like orthogonality or when expressing the equation of a circle.

Transcript

00:02 Hi, we're giving here two points on the circle, a1, the point as coordinate as v1 and a2, z2.

00:11 There are just two points, end points of a diameter.

00:14 That is given to us.

00:16 Statement 1, so it says that this is the equation of a circle.

00:20 The statement 2 we have.

00:22 Statement 2 says that if you take any point here, z and a point of z in the circle, then we get this equation.

00:32 Check here statement 1 and statement 2.

00:35 So if we join here, we'll see with a, we know a one into the diameter, the join this we have.

00:46 So we join this, it's angle to come out of 19 here.

00:50 This is 90.

00:52 So what you can do now, we can use here the pythagoras theorem.

00:59 So as we use it, so we get the first.

01:04 This here is given as more z negative z, z, 1.

01:09 So we have mod v negative z1 square in the pythagoras theorem here.

01:16 Then we have more v negative z2.

01:24 That equals we'll get more z1 negative z2 square by the other theorem.

01:35 I will just expand it what's coming out to be more z square.

01:41 Next we have positive mod z1 square.

01:45 Next the even conjugate next z conjugate z plus mod z squared positive mode z2 square negative z v2 conjugate negative z negative z conjugate z that's given as equal to more we have v1 square plus mode v2 square negative z2 conjugate negative v1 conjugate z2 so what's we have got here now this will cancel now let's see anything more cancels here then we have z1 z2 conjugate so we can just rewrite that so what's coming out to be we get two times mod z square negative z z z z conjugate negative z conjugate z one negative z two conjugate negative z two negative z conjugate z two and we have positive v1 z2 conjugate positive even conjugate z1 conjugate z that equal zero we get this equation now in the pythagoras theorem now if we just look at statement two here let's work on it for it given as z negative z 1 times z conjugate negative z 2 conjugate plus z conjugate negative z 1 conjugate times z negative z 2 that equal 0 now let's expand this so we get from this z z z conjugate negative v v2 conjugate negative z 1 z conjugate positive z 1 2 conjugate then we have positive z conjugate z negative z 2 z conjugate negative z 2 2nd conjugate z and we have positive z 1 conjugate z 2 that equals 0 let's compare here this equation and this equation here we have so we have 2 more z squared so from here z z conjugate and this we get 2 more z square so it's forgetting next we have negative z z z z one conjugate so here we have negative z z z even conjugate next negative z conjugate z one then negative z conjugate g one next we have negative z z two conjugate maybe have next negative z conjugate z two maybe have then positive z one z two conjugate maybe half then probably is even conjugate z2 maybe have so the two equations are equal so we can say that the statement two is coming out to be correct and let's look at statement one so let's look on statement one now statement one we take here model 2 z negative z1 negative z2 equals model z1 negative z2 this square both sides and we use that the complex number multiple that is a conjugate...

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