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COMPLEX NUMBERS AND QUADRATIC EQUATIONS — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSCOMPLEX NUMBERS AND QUADRATIC EQUATIONS3 Marks
Question
Find the modulus and argument of the complex number $\frac{{1 + 2i}}{{1 - 3i}}$.
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Answer

Let $z = \frac{{1 + 2i}}{{1 - 3i}} \times \frac{{1 + 3i}}{{1 + 3i}} = \frac{{1 + 3i + 2i + 6{i^2}}}{{1 - 9{i^2}}}$
$ = \frac{{ - 5 + 5i}}{{10}} = \frac{{ - 1 + i}}{2}$
Now $z = \frac{{ - 1}}{2} + \frac{i}{2} = r(\cos \theta + i\sin \theta )$
$\therefore r\;\cos \theta = \frac{{ - 1}}{2}$ and $r\sin \theta = \frac{1}{2}$
Squaring both sides of (i) and adding
${r^2}({\cos ^2}\theta + {\sin ^2}\theta ) = \frac{1}{4} + \frac{1}{4}$$ \Rightarrow {r^2} = \frac{1}{2} \Rightarrow r = \frac{1}{{\sqrt 2 }}$
$\therefore \frac{1}{{\sqrt 2 }}\cos \theta = \frac{{ - 1}}{2}$ and $\frac{1}{{\sqrt 2 }}\sin \theta = \frac{1}{2}$
$ \Rightarrow \cos \theta = \frac{{ - 1}}{{\sqrt 2 }}$ and $\sin \theta = \frac{1}{{\sqrt 2 }}$
Since $\sin \theta $ is positive and $\cos \theta $ is negative
$\therefore \theta $ lies in second quadrant
$\therefore \theta = \left( {\pi - \frac{\pi }{4}} \right) = \frac{{3\pi }}{4}$
$\therefore $ Modulus of $z = \frac{1}{{\sqrt 2 }}$ and argument of $z = \frac{{3\pi }}{4}$.

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