Question

The $a + ib$ greater than $c + id$ can be explained only when
A. $b = 0,c = 0$
B. $b = 0,d = 0$
C. $a = 0,c = 0$
D. $a = 0,d = 0$

Answer 1

By the presence of $i$, in the question we can say that the given question is based on the complex numbers. A complex number is the sum of a real number and an imaginary number in which the imaginary number is represented with $i$. In this question we need to find the value of variables to make $a + ib > c + id$. Let us see this in detail.

Complete step by step answer:
The equality of complex numbers states that, the sum of two complex numbers is equal when their real parts and imaginary parts are equal: $a + ib = c + id \Rightarrow a = c,b = d$. But we are asked to find $a + ib > c + id$, which is an inequality where the inequality of the complex numbers is not defined.

Complex numbers have an imaginary part $i$, thus we cannot compare complex parts but their real numbers can be compared, that is, only in the absence of imaginary parts we can make a comparison. In mathematics the absence of a term can be done by multiplying the term with $0$, thus by making the coefficients of imaginary part $i$ as $0$, we can compare.

Here the coefficients of imaginary part is $b$ and $d$ thus by applying $b = 0,d = 0$, the imaginary part will become zero (any number multiplied by $0$ will be $0$), then $a + ib > c + id$ can be defined. $a + ib > c + id$ is meaningful if both the imaginary parts are zero and $a > c$. If $b = 0,d = 0$ then, $a + i0$.
$a + ib \Rightarrow a + i0$ and $c + id \Rightarrow c + i0$

Hence option B is correct.

Note: A number of the form $a + ib$, where $a$ and $b$ are real numbers, is called a complex number, $a$ is called the real part and $b$ is called the imaginary part of the complex number. Let ${z_1} = a + ib$ and ${z_2} = c + id$. Then $a + ib > c + id$ if and only if $b = 0,d = 0$ and $a > c$.