Argand plane and polar representation - class-XII

The equation arg((z-z1)/(z-z2)) = pi/4 represents an arc of a circle passing through z1 and z2. Given z1=8+4i and z2=6+4i, the midpoint is 7+4i. The geometry of the angle pi/4 implies a specific circle equation.

Let $z=4-2i$

Since both the real and imaginary parts are positive , 4 + j3 lies in first quadrant of argand/complex plane.

$Let\quad { z } _{ 1 }=2r{ e }^{ i{ \theta  } _{ 1 } },{ z } _{ 2 }=r{ e }^{ i{ \theta  } _{ 2 } }\ arg\left( { z } _{ 1 }{ z } _{ 2 } \right) ={ \theta  } _{ 1 }+{ \theta  } _{ 2 }=\cfrac { 3\pi  }{ 2 } \ \bar { { z } _{ 1 } } =2r{ e }^{ -i{ \theta  } _{ 1 } }\ \cfrac { \bar { { z } _{ 1 } }  }{ { z } _{ 2 } } =\cfrac { 2r{ e }^{ -i{ \theta  } _{ 1 } } }{ r{ e }^{ i{ \theta  } _{ 2 } } } =2r{ e }^{ -i({ \theta  } _{ 1 }+{ \theta  } _{ 2 }) }=2r{ e }^{ -i\cfrac { 3\pi  }{ 2 }  }=2i$

Given that

|z-4| < |z-2| represents the set of points closer to 4 than to 2 on the complex plane. This is the half-plane to the right of the perpendicular bisector of the segment connecting 2 and 4, which is the line Re(z) = 3.

$|e^{a + ib}| = |e^a . e^{ib}|$

Let $z=x+ iy$

Since z1 and z2 are roots of z^2 + pz + q = 0, we have z1 + z2 = -p and z1*z2 = q. Given OA = OB, the magnitudes |z1| = |z2|, meaning |z1|^2 = |z2|^2 = q. Using the geometric relation for the angle alpha between z1 and z2 from the origin, |z1 - z2|^2 = |z1|^2 + |z2|^2 - 2|z1||z2|cos(alpha) = 2q - 2q cos(alpha). Also, (z1 - z2)^2 = (z1 + z2)^2 - 4z1z2 = p^2 - 4q. Equating and simplifying gives p^2 = 4q cos^2(alpha/2).

This is a known identity in complex geometry for points on a circle centered at the origin. The sum of the vectors weighted by the sine of the angles relates to the geometry of the triangle inscribed in the circle.

Z = sin(6pi/5) + i(1 + cos(6pi/5)). Using half-angle identities: sin(6pi/5) = 2sin(3pi/5)cos(3pi/5) and 1+cos(6pi/5) = 2cos^2(3pi/5). Factoring out 2cos(3pi/5) gives Z = 2cos(3pi/5) * (sin(3pi/5) + i cos(3pi/5)). The argument is 9pi/10.

Fact:  $\overline z$ is image of $z$ in $x$-axis

$(1-i)z _1=2z _2$

Let $Z=re^{i\theta _1}$

Consider the given equation $(z+1)^4=16z^4$
$ \Rightarrow (z+1)^4=16z^4$

$ \Rightarrow z+1=(2^4z^4)^{\frac{1}{4}}$

$ \Rightarrow |z+1|=2|z|$

We know that $z=x+iy$

Therefore $|x+iy+1|=2|x+iy|$

$ \Rightarrow \sqrt{(x+1)^2+y^2}=2\sqrt{x^2+y^2}$

$ \Rightarrow (x+1)^2+y^2=4(x^2+y^2)$

$ \Rightarrow x^2+2x+1+y^2=4x^2+4y^2$

$ \Rightarrow 3x^2+3y^2-2x-1=0$

Divide throughout by 3 we get,
$ \Rightarrow x^2+y^2-\dfrac{2}{3}x-\dfrac{1}{3}=0$, which represents a circle.

We know that for the circle equation of the form $x^2+y^2+2gx+2hy+c=0$ the center of the circle is given by $(-g,-h)$

We have $3x^2+3y^2-2x-1=0$ where $g=-\dfrac{1}{3}, h=0$.

Hence the center is $(\dfrac{1}{3},0)$ which is equidistant from the root of the equation.