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Rewrite 4+i over 3-2i in the form a+bi, where a and b are rational numbers. please show steps, reviewing for exams and have no clue how to do.

Sagot :

superkoopasirf
[tex]\frac{4 + i}{3 - 2i} \\\\ = \frac{(4 + i)(3 + 2i)}{(3 - 2i)(3 + 2i)} \text{ (rationalising the denominator) } \\\\ = \frac{12 + 3i + 8i + 2i^2}{9 - 6i + 6i - 4i^2} \\\\ = \frac{12 + 11i + 2i^2}{9 - 4i^2} \\\\ = \frac{12 + 11i + 2(-1)}{9 - 4(-1)} \\\\ = \frac{10 + 11i}{13} \\\\ = \frac{10}{13} + \frac{11}{13}i[/tex]
[tex]Use:(a-b)(a+b)=a^2-b^2\\-------------------------\\\\\frac{4+i}{3-2i}=\frac{4+i}{3-2i}\cdot\frac{3+2i}{3+2i}=\frac{(4+i)(3+2i)}{(3-2i)(3+2i)}=\frac{4(3)+4(2i)+i(3)+i(2i)}{3^2-(2i)^2}\\\\=\frac{12+8i+3i+2i^2}{9-2^2i^2}=\frac{12+11i+2(-1)}{9-4(-1)}=\frac{12+11i-2}{9+4}=\frac{10+11i}{13}\\\\=\boxed{\frac{10}{13}+\frac{11}{13}i}\\\\z=a+bi\Rightarrow a=\frac{10}{13}\ and\ b=\frac{11}{13}[/tex]
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