Question

Determine two complex numbers, (a+bi) and (c+di), where a and d are irrational numbers and b and c are rational number

Answer 1

Answer:

The two complex numbers are $\sqrt2 + 5i \text{ and } 6 + \sqrt5i$

Explanation:

We have to form two complex numbers of the form

$a + ib\\c + id$

such that and d are irrational numbers and b and c are rational numbers.

We know that $\sqrt2, \sqrt3$ are irrational numbers.

5 and 6 are rational numbers.

We put

$a = \sqrt2\\b = 5\\c = 6\\d = \sqrt5\\a+ib = \sqrt2 + 5i\\c + id = 6 + \sqrt5i$

Thus, the two complex numbers are: $\sqrt2 + 5i \text{ and } 6 + \sqrt5i$

Answer 2

Answer:

$a+ib=\sqrt{3}+2i$ and $c+id=3+\sqrt{2}i$.

Explanation:

Rational number: If a number is defined in the form of p/q, where p and q are integers and q≠0, then it is called a rational number.

For example: 2, 0.2, 3/4 etc.

Irrational number: If a number can not be defined in the form of p/q, where p and q are integers and q≠0, then it is called an irrational number.

For example: √2, 3.222.., π etc.

We need to find two complex numbers, (a+bi) and (c+di), where a and d are irrational numbers and b and c are rational number.

We can choose any irrational numbers for a and d.

$a=\sqrt{3},d=\sqrt{2}$

We can choose any rational numbers for b and c.

$b=2,c=3$

Two complex numbers are

$a+ib=\sqrt{3}+2i$

$c+id=3+\sqrt{2}i$

Therefore, the two complex numbers are $a+ib=\sqrt{3}+2i$ and $c+id=3+\sqrt{2}i$.