Question

Solve given question​

Answer 1

Answer:

Option (B) is correct

Step-by-step explanation:

We are given the complex number,

$\longrightarrow z=\dfrac{1 + 2i}{1 - (1 - i)^{2} }$

$\longrightarrow z=\dfrac{1 + 2i}{1 - (1^{2} + {i}^{2} - 2(1)(i) )}$

${\longrightarrow z=\dfrac{1 + 2i}{1 - (1 - 1 - 2i)}\qquad(\therefore i^2=-1)}$

$\longrightarrow z=\dfrac{1 + 2i}{1 - ( - 2i)}$

$\longrightarrow z=\dfrac{1 + 2i}{1 + 2i}$

$\longrightarrow z=1$

So the given complex number is 1.

We have to find it's amplitude and magnitude i.e. modulus.

So, modulus of complex number is given by,

$\leadsto |z| = \sqrt{Im(z)^2+Re(z)^2}$

$\leadsto |z| = \sqrt{(0)^2+(1)^2}$

$\leadsto |z| = \sqrt{0 + 1}$

$\leadsto |z| = 1$

So the modulus of the complex number is 1.

Now, the amplitude of complex number is given by,

${\implies Am(z) = \tan^-\left | \dfrac{Im(z)}{Re(z)}\right | }$

${\implies Am(z) = \tan^-\left | \dfrac{0}{1}\right | }$

${\implies Am(z) = \tan^-\left | 0\right | }$

${\implies Am(z) = 0}$

So the amplitude of complex number is 0.

Answer 2

$\large\underline{\sf{Solution-}}$

Given complex number is

$\rm :\longmapsto\:\dfrac{1 + 2i}{1 - {(1 - i)}^{2} }$

Let we assume that

$\rm :\longmapsto\:z = \dfrac{1 + 2i}{1 - {(1 - i)}^{2} }$

$\rm :\longmapsto\:z = \dfrac{1 + 2i}{1 - [1 + {i}^{2} - 2i] }$

$\rm :\longmapsto\:z = \dfrac{1 + 2i}{1 - [1 - 1 - 2i] }$

$\rm :\longmapsto\:z = \dfrac{1 + 2i}{1 - [- 2i] }$

$\rm :\longmapsto\:z = \dfrac{1 + 2i}{1 + 2i }$

$\bf\implies \:z = 1$

Now, let we assume that,

$\rm :\longmapsto\:1 = r(cos\theta + i \: sin\theta)$

$\rm :\longmapsto\:1 + 0i = rcos\theta + i \: r \: sin\theta$

On comparing, real and Imaginary parts, we get

$\rm :\longmapsto\:rcos\theta = 1 - - - (1)$

and

$\rm :\longmapsto\:rsin\theta = 0 - - - (2)$

On squaring equation (1) and (2), we get

$\rm :\longmapsto\: {r}^{2} {cos}^{2}\theta = 1$

and

$\rm :\longmapsto\: {r}^{2} {sin}^{2}\theta = 0$

On adding above 2 equations, we get

$\rm :\longmapsto\: {r}^{2} {cos}^{2}\theta + {r}^{2} {sin}^{2}\theta = 1 + 0$

$\rm :\longmapsto\: {r}^{2} ({cos}^{2}\theta + {sin}^{2}\theta )= 1$

$\rm :\longmapsto\: {r}^{2} = 1$

$\bf\implies \:r = 1$

On dividing equation (2) by equation (1), we get

$\rm :\longmapsto\:\dfrac{rsin\theta}{rcos\theta} = \dfrac{0}{1}$

$\rm :\longmapsto\:tan\theta = 0$

$\bf\implies \:\theta = 0$

Hence,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{ |z| = 1} \\ \\ &\sf{\theta = 0} \end{cases}\end{gathered}\end{gathered}$

Hence,

• Option (b) is correct.