Geometric representation of a complex number - 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.

The problem involves complex number transformations. Given the starting point and movements, the final argument calculation leads to an angle of 90 degrees, which corresponds to cot inverse of 0.

For any non-zero complex number z, -|z| is a negative real number. The argument of any negative real number is pi. Therefore, arg(-|z|) = pi. Integrating this constant function with respect to x from 0 to 100 gives the integral of pi dx from 0 to 100, which evaluates to pi * (100 - 0) = 100pi.

To find the quadrant, multiply the numerator and denominator by the conjugate (1+i). (1+2i)(1+i) / (1-i)(1+i) = (1+i+2i-2) / 2 = (-1+3i) / 2 = -0.5 + 1.5i. This point lies in the second quadrant.

Let z = x + iy. Then z^2 = (x^2 - y^2) + i(2xy), so |z^2| = x^2 + y^2 and |z^2 - 1| = sqrt((x^2 - y^2 - 1)^2 + (2xy)^2). The given equation becomes |z^2 - 1| = |z^2| + 1. Squaring both sides yields (x^2 - y^2 - 1)^2 + 4x^2y^2 = (x^2 + y^2 + 1)^2, which simplifies to x^2 = 0, meaning x = 0. Thus, z lies on the imaginary axis.