Complex numbers

[TsAmE]

Find integers p and q such that $\displaystyle (3 + 7i)(p + qi)$ is purely imaginary.

Attempt:

$\displaystyle (3 + 7i)(p + qi)$
$\displaystyle =3p + 3qi + 7pi - 7q$
$\displaystyle =3p - 7q + 3qi + 7pi$

Since 3p - 7q is the real part, it must = 0 therefore:

$\displaystyle 3p - 7q = 0$

$\displaystyle p = \frac{7q}{3}$

and

$\displaystyle q = \frac{3p}{7}$

but the correct answer was p = 7n and q = 3n, where nEZ. I am confused :?

 

[tkhunny]

That's what you have. Where is your confusion?

3p-7q = 0

If n is a complex number, then p = 7n and q = 3n, giving: 3(7n) - 7(3n) = 21n - 21n = 0

 

[TsAmE]

That's what you have. Where is your confusion?

3p-7q = 0

If n is a complex number, then p = 7n and q = 3n, giving: 3(7n) - 7(3n) = 21n - 21n = 0

I am not sure where you got p = 7n and q = 3n using 3p - 7q = 0

 

[lookagain]

Find integers p and q such that $\displaystyle (3 + 7i)(p + qi)$ is purely imaginary.

$\displaystyle 3p - 7q = 0$

$\displaystyle p = \frac{7q}{3}$

and

$\displaystyle q = \frac{3p}{7}$

but the correct answer was p = 7n and q = 3n, where nEZ. I am confused :?

When you get it to $\displaystyle 3p - 7q = 0,$ then you can change it to:

$\displaystyle 3p = 7q$

$\displaystyle \frac{3p}{q} = \frac{7q}{q}$

$\displaystyle \frac{3p}{q} = 7$

$\displaystyle \frac{1}{3}( \frac{3p}{q}) = \frac{1}{3}(\frac{7}{1})$

$\displaystyle \frac{p}{q} = \frac{7}{3}$

$\displaystyle \frac{7}{3} = \frac{7 \cdot 1 }{3 \cdot 1} = \frac{7 \cdot 2}{3 \cdot 2} = \frac{7 \cdot 3}{3 \cdot 3} = \ . . . = \frac{7 \cdot (-1)}{3 \cdot (-1)} = \frac {7 \cdot (-2)}{3 \cdot (-2)} = \ . . .$


The numerator can be any integer multiple of 7, and the denominator can be any corresponding integer multiple of 3,

$\displaystyle except \ a \ zero \ multiple, \ because \ \ \frac {7 \cdot 0}{3 \cdot 0} = \frac {0}{0} \ \ is \ undefined.$

Therefore, $\displaystyle \frac{p}{q} = \frac{7n}{3n},$ where n is a nonzero integer.

So, taking apart this fraction, $\displaystyle \ p = 7n \ and \ q = 3n.$