Gujarat BoardEnglish MediumSTD 11 ScienceMATHSComplex Numbers4 Marks
Question
Express the following complex numbers in the form $\text{r}(\cos\theta+\text{i}\sin\theta):$ $1-\sin\alpha+\text{i}\cos\alpha$
✓
Answer
Let $\text{z}=(1-\sin\alpha)+\text{i}\cos\alpha$ Since sine and cosine are periodic functions with periodic with peroid $2\pi.$ So, let us take $\alpha$ lying in the interval $[0,2\pi]$ Now, $\text{z}=(1-\sin\alpha)+\text{i}\cos\alpha$ $\Rightarrow|\text{z}|=\sqrt{(1-\sin\alpha)^2+\cos^2\alpha}\\=\sqrt{2-2\sin\alpha}=\sqrt{2}\sqrt{1-\sin\alpha}$ $\Rightarrow|\text{z}|=\sqrt{2}\sqrt{\Big(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big)^2}=\sqrt{2}\Big|\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big|$ Let $\beta$ be acute angle given by $\tan\beta=\frac{|\text{Im(z)}|}{|\text{Re(z)|}}.$ $\tan\beta=\frac{|\cos\alpha|}{|1-\sin\alpha|}=\frac{|\cos\alpha|}{|1-\sin\alpha|}=\Bigg|\frac{\cos^2\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}}{\Big(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big)}\Bigg|=\Bigg|\frac{\cos\frac{\alpha}{2}+\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}}\Bigg|$ $\Rightarrow\tan\beta=\Bigg|\frac{1+\tan\frac{\alpha}{2}}{1-\tan\frac{\alpha}{2}}\Bigg|=\Big|\tan\big(\frac{\pi}{4}+\frac{\alpha}{2}\big)\Big|$ Following cases arise: Case I: when $0\leq\alpha<\frac{\pi}{2}$ $\cos\frac{\alpha}{2}>\sin\frac{\alpha}{2}$ and $\frac{\pi}{4}+\frac{\alpha}{2}\in\Big[\frac{\pi}{4},\frac{\pi}{2}\Big)$ $\therefore\text{arg(z)}=\frac{\pi}{2}+\frac{\alpha}{2}$ So polar form of z is $\sqrt{2}\Big(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big)\Big(\cos\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)+\text{i}\sin\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)\Big)$ Case II: when $\frac{\pi}{2}<\alpha<\frac{3\pi}{2}$ $\cos\frac{\alpha}{2}<\sin\frac{\alpha}{2}$ and $\frac{\pi}{4}+\frac{\alpha}{2}\in\Big(\frac{\pi}{2},\pi\Big)$ $\therefore|\text{z}|=\sqrt{2}\Big|\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big|=\sqrt{2}\Big(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big)$ and, $\tan\beta=\Bigg|\tan\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)\Bigg|=-\tan\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)\\=\tan\Big\{\pi\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)\Big\}=\tan\Big(\frac{3\pi}{4}-\frac{\alpha}{2}\Big)$ $\Rightarrow\beta=\frac{3\pi}{4}-\frac{\alpha}{2}$ Since $1-\sin\alpha>0$ and $\cos\alpha<0.$ Clearly, z lies in the fourth quadrant. $\therefore\text{arg(z)}=-\beta=\frac{\alpha}{2}-\frac{3\pi}{4}$ So polar form of z is $\sqrt{2}\Big(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big)\Big(\cos\Big(\frac{\alpha}{2}-\frac{3\pi}{4}\Big)+\text{i}\sin\Big(\frac{\alpha}{2}-\frac{3\pi}{4}\Big)\Big)$ Case III: when $\frac{3\pi}{2}<\alpha<2\pi$ $\cos\frac{\alpha}{2}<\sin\frac{\alpha}{2}$ and $\frac{\pi}{4}+\frac{\alpha}{2}\in\Big(\pi,\frac{5\pi}{4}\Big)$ $\therefore|\text{z}|=\sqrt{2}\Big|\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big|=-\sqrt{2}\Big(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big)$ and, $\tan\beta=\Bigg|\tan\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)\Bigg|=\tan\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)\\=-\tan\Big\{\pi-\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)\Big\}=\tan\Big(\frac{\alpha}{2}-\frac{3\pi}{4}\Big)$ $\Rightarrow\beta=\frac{\alpha}{2}-\frac{3\pi}{4}$ Clearly, $\text{Re(z)}<0$ and $\text{Im(z)}>0.$ $\therefore\text{arg(z)}=\beta=\frac{\alpha}{2}-\frac{3\pi}{4}$ So polar form of z is $-\sqrt{2}\Big(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big)\Big(\cos\Big(\frac{\alpha}{2}-\frac{3\pi}{4}\Big)+\text{i}\sin\Big(\frac{\alpha}{2}-\frac{3\pi}{4}\Big)\Big).$
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