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

Hint: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 \[a > c\].