Question

If ${a+ib}<{c+id}$ is meaningful if:
(a) \[{{a}^{2}}+{{b}^{2}}=0\]
(b) \[{{b}^{2}}+{{c}^{2}}=0\]
(c) \[{{a}^{2}}+{{c}^{2}}=0\]
(d) \[{{b}^{2}}+{{d}^{2}}=0\]

Answer 1

Hint:In this question, we need to know the condition for comparing the complex numbers. Then as we get that they cannot be compared we need to make the imaginary part 0 and solve further to get the relation.

Complete step-by-step answer:
COMPLEX NUMBER:
A number of the form \[z=x+iy\], where x and y belongs to the real numbers, is called a complex number.
The numbers x and y are called respectively real and imaginary parts of a complex number z.
Purely Real and Purely Imaginary Complex Number:
A complex number z is purely real, if its imaginary part is 0. And purely imaginary, if its real part is 0.
To equate any two complex numbers we need to equate the corresponding real part and imaginary part. But, we cannot compare two complex numbers directly.
Now, from the given inequality in the question we get,
$\Rightarrow {a+ib}<{c+id}$
Here, as we cannot compare the complex numbers directly so in order to compare them they should be purely real which means that their imaginary parts should be zero.
Now, the imaginary parts in the given complex numbers are b and d
Now, to compare the given two imaginary numbers from the above condition we get,
\[\Rightarrow b=0,d=0\]
Now, this can also be written in the form of an equation as
\[\Rightarrow {{b}^{2}}+{{d}^{2}}=0\]
Hence, the correct option is (d).

Note:It is important to note that complex numbers cannot be compared as the imaginary part cannot be decided greater or lesser than other numbers. So, to compare the complex numbers the necessary condition is that they should be purely real but two complex numbers can be equated.Here, as b and d are 0 instead of directly writing the equation we can first express their sum as 0 and then by squaring on both sides to show that the sum of their squares is 0.