How do you divide $\frac{6 - 3 i}{2 + i}$ $(6-3i)/(2+i)$ ?

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Answer 1 · 2018-06-27T09:48:43

See explanation below

We will use the conjugate of complex number in denominator

$2 - i$ $2-i$ . We multiply and divide by this number, so the quotient does not varies

$\frac{(6 - 3 i) (2 - i)}{(2 + i) (2 - i)} = \frac{12 - 6 i - 6 i + 3 i^{2}}{4 - i^{2}} =$ $((6-3i)(2-i))/((2+i)(2-i))=(12-6i-6i+3i^2)/(4-i^2)=$

But we know that $i^{2} = - 1$ $i^2=-1$ , then

$\frac{9 - 12 i}{4 + 1} = \frac{9}{5} - \frac{12}{5} i$ $(9-12i)/(4+1)=9/5-12/5i$

How do you divide (6-3i)/(2+i)6−3i2+i(6-3i)/(2+i)?