Statement (a + ib) < (c + id) is true for

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4-1.Complex numbers

normal

The statement $(a + ib) < (c + id)$ is true for

A

${a^2} + {b^2} = 0$

B

${b^2} + {c^2} = 0$

C

${a^2} + {c^2} = 0$

D

${b^2} + {d^2} = 0$

Solution

(d) $a + ib < c + id,\,$ defined if and only if its imaginary parts must be equal to zero, i.e. $b = d = 0.$

So,${b^2} + {d^2} = 0$.
.

Standard 11

Mathematics

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