### Compare the real parts and imaginary parts of two imaginary numbers using rationalization

Question Sample Titled 'Compare the real parts and imaginary parts of two imaginary numbers using rationalization'

Let ${u}=\dfrac{{3}}{{{a}+{5}{i}}}$ and ${v}=\dfrac{{3}}{{{a}-{5}{i}}}$ , where ${a}$ is a real number. Which of the following must be true?

 I. ${u}{v}$ is a rational number.  II. The real part of ${u}$ is equal to the real part of ${v}$ . III. The imaginary part of $\dfrac{{1}}{{u}}$ does not equal to the imaginary part of $\dfrac{{1}}{{v}}$ .

A
II and III only
B
II only
C
III only
D
I and II only

 I.  ${u}{v}$ $=\dfrac{{3}}{{{a}+{5}{i}}}\times\dfrac{{3}}{{{a}-{5}{i}}}$ $=\dfrac{{9}}{{{\left({a}+{5}{i}\right)}{\left({a}-{5}{i}\right)}}}$ $=\dfrac{{9}}{{{a}^{{2}}-{25}{i}^{{2}}}}$ $=\dfrac{{9}}{{{a}^{{2}}+{25}}}$ ${i}^{{2}}=-{1}$ Take ${a}$ as some irrational number whose square is still irrational, i.e. $\pi,{2}^{{\tfrac{{1}}{{3}}}},{\log{{3}}}$ etc. ${a}^{{2}}+{25}$ is not rational. So the whole fraction is irrational. I is false. II.  ${u}=\dfrac{{3}}{{{a}+{5}{i}}}=\dfrac{{{3}{\left({a}-{5}{i}\right)}}}{{{\left({a}+{5}{i}\right)}{\left({a}-{5}{i}\right)}}}=\dfrac{{{3}{a}-{15}{i}}}{{{a}^{{2}}+{25}}}$ $=\dfrac{{{3}{a}}}{{{a}^{{2}}+{25}}}+{\left(\dfrac{{-{15}}}{{{a}^{{2}}+{25}}}\right)}{i}$ Similarly,  ${v}=\dfrac{{3}}{{{a}-{5}{i}}}=\dfrac{{{3}{\left({a}+{5}{i}\right)}}}{{{\left({a}-{5}{i}\right)}{\left({a}+{5}{i}\right)}}}=\dfrac{{{3}{a}+{15}{i}}}{{{a}^{{2}}+{25}}}$ $=\dfrac{{{3}{a}}}{{{a}^{{2}}+{25}}}+{\left(\dfrac{{15}}{{{a}^{{2}}+{25}}}\right)}{i}$ Notice that ${a}$ is a real number,  ∴   The real part of ${u}=$the real part of ${v}=\dfrac{{{3}{a}}}{{{a}^{{2}}+{25}}}$ II is true. III.  $\dfrac{{1}}{{u}}$ $=\dfrac{{{a}+{5}{i}}}{{3}}=\dfrac{{a}}{{3}}+\dfrac{{5}}{{3}}{i}$ $\dfrac{{1}}{{v}}$ $=\dfrac{{{a}-{5}{i}}}{{3}}=\dfrac{{a}}{{3}}-\dfrac{{5}}{{3}}{i}$ The imaginary part of ${u}$ $=\dfrac{{5}}{{3}}$ The imaginary part of ${v}$ $=-\dfrac{{5}}{{3}}$ III is true.

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