Solve the Following Question.(5 Marks)

Question 15 Marks

What is the smallest positive integer n for which $(1+\text{i})^{2\text{n}}=(1-\text{i})^{2\text{n}} \ ?$

Answer

$(1+\text{i})^{2\text{n}}=(1-\text{i})^{2\text{n}}$
$\Rightarrow\Big(\frac{1+\text{i}}{1-\text{i}}\Big)^{2\text{n}}=1$
$\Rightarrow\Big(\frac{(1+\text{i})(1+\text{i})}{(1-\text{i})(1+\text{i})}\Big)^{2\text{n}}=1$ [Rationalizing the denominator]
$\Rightarrow\Big(\frac{1+2\text{i}-1}{1+1}\Big)^{2\text{n}}=1$
$\Rightarrow\Big(\frac{2\text{i}}{2}\Big)^{2\text{n}}=1$
$\Rightarrow\text{i}^{2\text{n}}=1$
$\therefore\text{n}=2$

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Question 25 Marks

If $\Big(\frac{1-\text{i}}{1+\text{i}}\Big)^{100}=\text{a}+\text{ib},$ find (a, b).

Answer

$\Big(\frac{1-\text{i}}{1+\text{i}}\Big)^{100}=\text{a}+\text{ib}$
$\Rightarrow\Big(\frac{(1-\text{i})(1-\text{i})}{(1+\text{i})(1-\text{i})}\Big)^{100}=\text{a}+\text{ib}$ [Rationalizing the denominator]
$\Rightarrow\Big(\frac{(1-2\text{i}-1)}{(1+1)}\Big)^{100}=\text{a}+\text{ib}$
$\Rightarrow\Big(\frac{-2\text{i}}{2}\Big)^{100}=\text{a}+\text{ib}$
$\Rightarrow(-\text{i})^{100}=\text{a}+\text{ib}$
$\Rightarrow1=\text{a}+\text{ib}$
Comparing, we get
$(\text{a},\text{b})=(1,0)$

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Question 35 Marks

Solve the equation $|\text{z}|=\text{z}+1+2\text{i}.$

Answer

Let $\text{z}=\text{x}+\text{iy}.$
Then,
$|\text{z}|=\sqrt{\text{x}^2+\text{y}^2}$
$\therefore|\text{z}|=\text{z}+1+2\text{i}$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=(\text{x}+\text{iy})+1+2\text{i}$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=(\text{x}+1)+\text{i}(\text{y}+2)$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=(\text{x}+1)$ and $\text{y}+2=0$
${\text{x}^2+\text{y}^2}=(\text{x}+1)^2$ and $\text{y}=-2$
${\text{x}^2+\text{y}^2}=\text{x}^2+1+2\text{x}$ and $\text{y}=-2$
$\Rightarrow\text{y}^2=2\text{x}+1$ and $\text{y}=-2$
$\Rightarrow4=2\text{x}+1$ and $\text{y}=-2$
$\Rightarrow2\text{x}=3$ and $\text{y}=-2$
$\Rightarrow\text{x}=\frac{3}{2}$ and $\text{y}=-2$
$\therefore\text{z}=\text{x}+\text{iy}=\frac{3}{2}-2\text{i}$

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Question 45 Marks

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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Maths Solve the Following Question.(5 Marks)

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