For a C, let A=z C:Re(a+z)Im(a+z) and B=z C:Re(a+z)(S1): If Re(a),Im(a)0, then the set A…

Question

For $a\in C$, let $A={z\in C:Re(a+\vec{z})>Im(\vec{a}+z)}$ and $B=z\in C:Re(a+\vec{z})(S1): If $Re(a),Im(a)>0$, then the set A contains all the real numbers
(S2) : If $

Accepted Answer

We are given that $a\in C$ and $z\in C$. Let $a={x}_{1}+i{y}_{1}$ and $z={x}_{2}+i{y}_{2}$ where ${x}_{1},{y}_{1},{x}_{2},{y}_{2}\in R$ 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 $(a+\vec{z})$ is greater than the imaginary part of $(\vec{a}+z)$. - B is the set of all complex numbers $z$ for which the real part of $(a+\vec{z})$ is less than the imaginary part of $(\vec{a}+z)$. 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 $z\in A$, $Re(a+\vec{z})>Im(\vec{a}+z)$ 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+\vec{z})

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