I have attempted this complex number below. Is it correct?
$$\frac {4 + i1} {2 + i3} $$
This is what i have:
$$ \frac {4 + i1} {2 + i3} \times \frac {2 + i3} {2 + i3} $$
$$ \frac {8-12i +2 -3i^2} {4 -6i + 6 - 9i^2} $$
$$ \frac {8 -12i +2 -3i^2 (-1)} {4 - 6i + 6 -9i^2}$$
$$ \frac {8 -12i +2 + 31)} {4 - 6i + 6 + 9}$$
$$ \frac {39 - 10i} {13} $$
Is this correct??
$\endgroup$ 42 Answers
$\begingroup$No, and that is not the simplest approach. You can do it as follows:\begin{align}\frac{4+i}{2+3i}&=\frac{(4+i)(2-3i)}{(2+3i)(2-3i)}\\&=\frac{11-10i}{13}\\&=\frac{11}{13}-\frac{10}{13}i.\end{align}
$\endgroup$ 4 $\begingroup$No. Use the opposite sign for the imaginary part in the denominator:
$$\frac {4 + 1i} {2 + 3i} = \frac {4 + 1i} {2 + 3i}\cdot \frac {2 - 3i} {2 - 3i}$$
to may use - in the denominator - the formula$$ (A+iB).(A-iB) = A^2 + B^2$$
and obtain (still in the denominator) a real number.
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