How do you perform the operation and write the result in standard form given (8-i)-(4-i)?

Sep 8, 2017

Let the complex number in standard form be,

$z = a + i b$ where $a$ and $b$ are real numbers.

Thus, for $z = a + i b = \left(8 - i\right) - \left(4 - i\right)$
$\implies a + i b = 8 - i - 4 + i$
$\implies a + i b = 4$

Comparing both sides,

$a = 4$ and $b = 0$

Therefore, $z = 4 + 0 \cdot i$ in standard form.