Question
Question: The \[a + ib\] greater than \[c + id\] can be explained only when A. \[b = 0,c = 0\] B. \[b = 0...
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
Solution
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⇒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⇒a+i0 and c+id⇒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 z1=a+ib and z2=c+id. Then a+ib>c+id if and only if b=0,d=0 and a>c.