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

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

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Answer 1 · 2017-09-08T15:35:03

Let the complex number in standard form be,

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

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

Comparing both sides,

$a = 4$ $a = 4$ and $b = 0$ $b = 0$

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

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

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