$\frac{8−i}{3−2i}$
If the expression above is rewritten in the form a+bi, where a and b are real numbers, what is the value of a?
All I know is that it equals $\frac{8-i}{3-2i}$ times $\frac{3+2i}{3+2i}$ and then i did not know what to do
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Show please your attempts. – Michael Rozenberg Mar 21 '20 at 02:21
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Well that would mean it is $\frac{(8-i)(3+2i)}{(3-2i)(3+2i)}$. And you ought to know that $(A+B)(C+D) = AC +AD +BC +BD$. So you should be able to expand the numerator and denominator out. There's a "trick" that $i^2 = -1$ but other than that.... it's just like any other multiplication. – fleablood Mar 21 '20 at 02:36
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Note,
$$\frac{8−i}{3−2i}=\frac{8-i}{3-2i}\cdot\frac{3+2i}{3+2i}=\frac{(24+2)+(16-3)i}{3^2-(2i)^2}=\frac{26+13i}{13}=2+i$$
Quanto
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Yes, but how did you get the bracketed fraction? That is what I was struggling on. To simplify the two fractions. Any way, can you tell me? – Stevo Mar 21 '20 at 02:33
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$(8-i)(3+2i)$ will be $24+16i-3i+2$ because $i×i=-1$ And $(3-2i)(3+2i)$ [using the identity $(x^2-y^2)=(x-y)(x+y)$] will be $9+4=13$ and then you can separate the fractions with the denominator $13$ and you will get the value of a which will be $26/13=2$.
Mansi Tyagi
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