Gujarat BoardEnglish MediumSTD 11 ScienceMATHSCOMPLEX NUMBERS AND QUADRATIC EQUATIONS4 Marks
Question
Find the modulus and argument of the complex number $\frac{1+2\text{i}}{1-3\text{i}}.$
✓
Answer
Let $\text{z}=\frac{1+2\text{i}}{1-3\text{i}},$ then $\text{z}=\frac{1+2\text{i}}{1-3\text{i}}\times\frac{1+3\text{i}}{1+3\text{i}}=\frac{1+3\text{i}+{2\text{i}+6\text{i}^2}}{1^2+3^2}=\frac{1+5\text{i}+6(-1)}{1+9}$ $=\frac{-5+5\text{i}}{10}=\frac{-5}{10}+\frac{5\text{i}}{10}=\frac{-1}{2}+\frac{1}{2}\text{i}$ Let $\text{z}=\text{r}\cos\theta+\text{ir}\sin\theta$ i.e., $\text{r}\cos\theta=\frac{-1}{2} \ \text{and r}\sin\theta=\frac{1}{2}$ On squaring and adding, we obtain $\text{r}^2(\cos^2\theta+\sin^2\theta)=\Big(\frac{-1}{2}\Big)^2+\Big(\frac{1}{2}\Big)^2$ $\Rightarrow\text{r}^2=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$ $\therefore\ \frac{1}{\sqrt{2}}\cos\theta =\frac{-1}{2}\text{ and }\frac{1}{\sqrt{2}}\sin\theta=\frac{1}{2}$ $\Rightarrow\ \cos\theta =\frac{-1}{\sqrt{2}}\text{ and }\sin\theta=\frac{1}{\sqrt{2}}$ $\therefore\ \theta=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$ [As $\theta$ lies in the II quadrant]Therefore, the modulus and argument of the given complex number are $\frac{1}{\sqrt{2}}$ and $\frac{3\pi}{4}$ respectively.
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