Complex Numbers

We are given that $a \in \mathbb{C}$ and $z \in \mathbb{C}$.

Let $a = x_1 + iy_1$ and $z = x_2 + iy_2$ where $x_1, y_1, x_2, y_2 \in \mathbb{R}$ We are also given two sets A and B defined as follows : - A is the set of all complex numbers $z$ for which the real part of $(a + \overline{z})$ is greater than the imaginary part of $(\overline{a} + z)$. - B is the set of all complex numbers $z$ for which the real part of $(a + \overline{z})$ is less than the imaginary part of $(\overline{a} + z)$.

Statement (S1) says : If the real part and imaginary part of $a$ are both positive, then the set A contains all the real numbers.

Statement (S2) says : If the real part and imaginary part of $a$ are both negative, then the set B contains all the real numbers.

We need to determine which of these statements are true.

Let's evaluate each statement.

1.

Statement (S1) : For $z \in A$, $Re(a + \overline{z}) > Im(\overline{a} + z)$ This can be re-written as $x_1 + x_2 > y_2 - y_1$ If we consider only real z (i.e. $y_2 = 0$) and given that $x_1, y_1 > 0$, then the condition simplifies to $x_2 > -(x_1 + y_1)$.

This indicates that A covers a part of the negative real axis, but not the entire real axis.

Therefore, Statement (S1) is false.

2.

Statement (S2) : For $z \in B$, $Re(a + \overline{z}) < Im(\overline{a} + z)$ This can be re-written as $x_1 + x_2 < y_2 - y_1$ If we consider only real z (i.e. $y_2 = 0$) and given that $x_1, y_1 < 0$, then the condition simplifies to $x_2 < -(x_1 + y_1)$.

This indicates that B covers a part of the positive real axis, but not the entire real axis.

Therefore, Statement (S2) is false.

Therefore, both (S1) and (S2) are false, so the answer is Option A : both are false.

To solve the problem, let's break it down step by step.

Step 1 : Find $w_{1}$ Given $z_{1} = 5 + 4i$.

When you rotate $z_{1}$ by $90^{\circ}$ anticlockwise about the origin, the real part becomes negative of the imaginary part of $z_{1}$, and the imaginary part becomes the real part of $z_{1}$.

Therefore, $w_{1}$ becomes : $w_{1} = -4 + 5i$ Step 2 : Find $w_{2}$ Given $z_{2} = 3 + 5i$.

When you rotate $z_{2}$ by $90^{\circ}$ clockwise about the origin, the real part becomes the imaginary part of $z_{2}$, and the imaginary part becomes negative of the real part of $z_{2}$.

Therefore, $w_{2}$ becomes : $w_{2} = 5 - 3i$ Step 3 : Calculate $w_{1} - w_{2}$ $w_{1} - w_{2} = (-4 + 5i) - (5 - 3i)$ $w_{1} - w_{2} = -9 + 8i$ Step 4 : Find the principal argument To find the principal argument, we need to compute the tangent inverse of the ratio of the imaginary part to the real part.

Argument = $\tan^{-1}\left(\dfrac{\text{Imaginary part}}{\text{Real part}}\right)$ Argument = $\tan^{-1}\left(\dfrac{8}{-9}\right)$ Argument = $-\tan^{-1}\left(\dfrac{8}{9}\right)$ Since it's in the third quadrant, the principal argument is : Argument = $\pi + (-\tan^{-1}\left(\dfrac{8}{9}\right))$ Argument = $\pi - \tan^{-1}\left(\dfrac{8}{9}\right)$ So, the correct option is : Option C : $\pi-\tan^{-1} \dfrac{8}{9}$

Given that $z=x+i y$

Since, $\dfrac{2 z-3 i}{4 z+2 i}$ is Real $\Rightarrow \operatorname{Img}\left(\dfrac{2 z-3 i}{4 z+2 i}\right)=0$

Also, $(4 x)^2+(4 y+2)^2 \neq 0 \Rightarrow y+x^2+y^2 \neq \dfrac{-1}{4}$ If $x=0$, then $y \neq-\dfrac{1}{2}$

Let, $z=x+i y$ So, $\dfrac{2 z-3 i}{2 z+i}$ is purely imaginary $~~$[Given]

Given that, $x+y^2=0$

Hence, required answer is $\dfrac{3}{4}$.

Then

$\because z$ is purely imaginary, so $\operatorname{Re} z=0$

Given : $|z+2|=z+4(1+i)$ Also, $z=\alpha+i \beta$

Let $z=x+i y$

$\begin{array}{ll}\operatorname{Re}\left(a z^2+b z\right)=b \\\\ \Rightarrow a\left(x^2-y^2\right)+b x=a \\\\ \Rightarrow x^2-y^2+\dfrac{b}{a} x=1 .........(i)\end{array}$ $$\begin{aligned} & \text{ and } b z^2+a z \\\\ & =b\left(x^2-y^2+2 i x y\right)+a(x+i y) \\\\ & =b\left(x^2-y^2\right)+a x+2 b i x y+a i y \\\\ & \operatorname{Re}\left(b z^2+a z\right)=b \\\\ & \Rightarrow b\left(x^2-y^2\right)+a x=b \\\\ & \Rightarrow x^2-y^2+\dfrac{a}{b} x=1 ..........(ii)\end{aligned}$$ On subtracting Equation (ii) from Equation (i), we get

From (i), when $x=0$, then

When, $a=-b$, then $x^2-y^2-x=1$ has infinitely many solutions.

To find the minimum value of $\left|z+\dfrac{1}{2}(3+4i)\right|$, where $z$ is a complex number with $|z| \leqslant 1$, we can think of this geometrically as the distance from any point inside or on the boundary of the unit circle in the complex plane to the fixed point $\dfrac{1}{2}(3+4i)$.

To make this more clear, first write the expression as follows:

This means we are looking for the distance between any point on the complex plane represented by $z$ (where $z$ has a magnitude of up to 1) and the point $1.5+2i$.

The triangle inequality gives us the following relationship for any two complex numbers $z_1$ and $z_2$:

Applying this to our specific case ($z_1 = z$ and $z_2 = -1.5 - 2i$):

Since $z$ is within the unit circle, $|z| \leqslant 1$ and $|-1.5 - 2i| = \sqrt{1.5^2 + 2^2} = \sqrt{2.25 + 4} = \sqrt{6.25} = 2.5$.

So now we have:

Which simplifies to:

Therefore, the minimum value of $\left|z+\dfrac{1}{2}(3+4i)\right|$ given that $|z| \leqslant 1$ is $\dfrac{3}{2}$.

So the correct answer is: Option C: $\dfrac{3}{2}$