Respuesta :
Answer:
[tex][(1+i\sqrt{29}), (1-i\sqrt{29})][/tex], [tex][(1+i\sqrt{26}), (1-i\sqrt{26})][/tex], [tex][(1+i\sqrt{21}), (1-i\sqrt{21})][/tex],
[tex][(1+i\sqrt{14}), (1-i\sqrt{14})][/tex], and [tex][(1+i\sqrt{5}), (1-i\sqrt{5})][/tex].
Step-by-step explanation:
Given that two complex numbers have a product of 30 and the real part of the complex numbers are positive integers.
Let one complex number is [tex]a+ib[/tex] where [tex]a[/tex] is a positive integer and [tex]b[/tex] is a real number.
As the product of two complex numbers is a purely real number, so the other complex number must be the conjugate of the first complex number.
So, another complex number [tex]= a-ib[/tex]
As a product of both the numbers = 30
[tex]\Rightarrow (a+ib)(a-ib)=30 \\\\\Rightarroe a^2 - (ib)^2=30 \\\\\Rightarrow a^2 - (i^2b^2)=30 \\\\Rightarrow a^2 - (-1\times b^2)= 30[/tex] [as i^2=-1]
[tex]\Rightarrow a^2+b^2= 30 \\\\\Rightarrow b^2 = 30-a^2 \\\\[/tex]
[tex]\Rightarrow b=\pm \sqrt {30-a^2} \cdots(i)[/tex]
Now, as a is a positive integer, so by taking possible values of a, we can determine the values of b from the equation (i).
For [tex]a=1[/tex]
[tex]b=\pm \sqrt {30-1^2}=\pm \sqrt {29}[/tex]
The positive sign is for the one number and the negative sign is for the conjugate number (another number).
So, the 1st combination is [tex][(1+i\sqrt{29}), (1-i\sqrt{29})][/tex]
Similarly,
For [tex]a=2[/tex]
[tex]b=\pm \sqrt {30-2^2}=\pm \sqrt {26}[/tex]
So, the 2nd combination is [tex][(1+i\sqrt{26}), (1-i\sqrt{26})][/tex]
For [tex]a=3[/tex]
[tex]b=\pm \sqrt {30-3^2}=\pm \sqrt {21}[/tex]
So, the 3rd combination is [tex][(1+i\sqrt{21}), (1-i\sqrt{21})][/tex]
For [tex]a=4[/tex]
[tex]b=\pm \sqrt {30-4^2}=\pm \sqrt {14}[/tex]
So, the 4th combination is [tex][(1+i\sqrt{14}), (1-i\sqrt{14})][/tex]
For [tex]a=5[/tex]
[tex]b=\pm \sqrt {30-5^2}=\pm \sqrt {5}[/tex]
So, the 5th combination is [tex][(1+i\sqrt{5}), (1-i\sqrt{5})][/tex]
The higher integral value of [tex]a[/tex] is not possible for any real [tex]b[/tex].
Hence, the possible combinations of required numbers are
[tex][(1+i\sqrt{29}), (1-i\sqrt{29})][/tex], [tex][(1+i\sqrt{26}), (1-i\sqrt{26})][/tex], [tex][(1+i\sqrt{21}), (1-i\sqrt{21})][/tex],
[tex][(1+i\sqrt{14}), (1-i\sqrt{14})][/tex], and [tex][(1+i\sqrt{5}), (1-i\sqrt{5})][/tex].
