Question

#Quality Question

@ Complex Numbers

Find all non zero complex number z satisfying
$\bar{z} = iz {}^{2}$
​

Answer 1

Answer

$z=i$ and $z=\pm \dfrac{\sqrt{3} }{2} -\dfrac{1}{2} i$.

Solution

Since $z$ is a complex number, let $z=a+bi$ $(a,b\in\mathbb{R},\ a,b\neq 0)$.

$\implies \bar{z}=iz^{2}$

$\implies a-bi=(a^{2}-b^{2})+2abi$

Since $a,b$ are real part and imaginary part respectively, we find a system equation with two unknowns:-

$\implies \begin{cases} & a=-2ab \ \ \ \ \ ...(1)\\ & -b=a^{2}-b^{2} \ \ \ \ \ ...(2)\end{cases}$

To solve this system equation we choose substitution method since the solutions of the first equation can be solved by factorization.

Solving $(1)$, we get,

$\implies a(2b+1)=0$

$\implies a=0\ \text{or}\ b=-\dfrac{1}{2}$

Using substitution method on $(2)$, we get,

$\implies a=0$

$\implies b^{2}-b=0$

$\implies b(b-1)=0$

$\implies b=0\text{(rej.)}\ \text{or}\ b=1$

or,

$\implies b=-\dfrac{1}{2}$

$\implies a^{2}=\dfrac{1}{4} +\dfrac{1}{2}$

$\implies a^{2}=\dfrac{3}{4}$

$\implies a=\pm \dfrac{\sqrt{3} }{2}$

⇒ The required answers are $z=i$ and $z=\pm \dfrac{\sqrt{3} }{2} -\dfrac{1}{2} i$.

Advanced problems

Question: Find the value of $\sqrt{i}$.

Answer: $\sqrt{i} =\pm\dfrac{\sqrt{2} }{2} (1+i)$

Answer key: Suppose $\sqrt{i}$ as a complex number.

Solution:-

Let $z=\sqrt{i}$. By definition,

$\implies z^{2}=i$

Suppose $z$ is a complex number, then

$\implies (a+bi)^{2}=i$

$\implies a^{2}-b^{2}+2abi=i$

We are left with a system equation,

$\implies \begin{cases} & a^{2}-b^{2}=0 \ \ \ \ \ ...(1)\\ & 2ab=1 \ \ \ \ \ ...(2)\end{cases}$

Solving $(1)$ we get,

$\implies (a+b)(a-b)=0$

$\implies a=\pm b$

Substitution method on $(2)$ gives,

$\implies a=b$

$\implies 2b^{2}=1$

$\implies b^{2}=\dfrac{1}{2}$

$\implies b=\pm\dfrac{\sqrt{2} }{2}$

$\implies (a,b)=(\pm \dfrac{\sqrt{2} }{2} ,\pm \dfrac{\sqrt{2} }{2} )$

or,

$\implies a=-b$

$\implies -2b^{2}=1>0$

Since $b$ is the imaginary part, the solution doesn't exist in this case.

Therefore the two square roots of $i$ are $\sqrt{i} =\pm \dfrac{\sqrt{2} }{2} (1+i)$.

Answer 2

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

Given equation is

$\rm :\longmapsto\:\bar{z} = i {z}^{2}$

On multiply both sides by 'i', we get

$\rm :\longmapsto\:i \: \bar{z} = {i}^{2} {z}^{2}$

$\rm :\longmapsto\:i \: \bar{z} = - {z}^{2} \: \: \: \: \: \: \: \purple {\{ \because \: {i}^{2} = - 1 \}}$

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

Since, z is a non zero complex number, so let assume that

$\rm :\longmapsto\:z = x + iy - - - (2)$

where x and y are non - zero real numbers.

So, on substituting the values, we get

$\rm :\longmapsto\: - \: i \: \: (\overline{x + iy}) = {(x + iy)}^{2}$

We know,

$\boxed{ \sf{ \:z = x + iy \: \: then \: \bar{z} = x - iy}}$

and

$\boxed{ \sf{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

So using this,

$\rm :\longmapsto\: - \: i(x - iy) = {x}^{2} + {i}^{2} {y}^{2} + 2xyi$

$\rm :\longmapsto\: - ix + {i}^{2}y = {x}^{2} - {y}^{2} + 2xyi$

$\rm :\longmapsto\: - ix - y = {x}^{2} - {y}^{2} + 2xyi$

$\rm :\longmapsto\: - y- ix = {x}^{2} - {y}^{2} + 2xyi - - - (3)$

On comparing Imaginary parts, we get

$\rm :\longmapsto\:2xy = - x$

$\rm :\longmapsto\:2xy + x = 0$

$\rm :\longmapsto\:x(2y + 1) = 0$

$\bf\implies \:y = - \dfrac{1}{2} \: \: or \: x \: = \: 0$

$\bf\implies \:y = - \dfrac{1}{2} - - - (4)$

Now, Comparing Real part on both sides, we get

$\rm :\longmapsto\: {x}^{2} - {y}^{2} = - y$

On substituting the value of y, we get

$\rm :\longmapsto\: {x}^{2} - \dfrac{1}{4} = \dfrac{1}{2}$

$\rm :\longmapsto\: {x}^{2} = \dfrac{1}{4} + \dfrac{1}{2}$

$\rm :\longmapsto\: {x}^{2} = \dfrac{1 + 2}{4}$

$\rm :\longmapsto\: {x}^{2} = \dfrac{3}{4}$

$\bf\implies \:x \: = \: \pm \: \sqrt{\dfrac{3}{4} }$

$\bf\implies \:x \: = \: \pm \: \dfrac{ \sqrt{3} }{2}$

And

when x = 0, we get y = 1 or y = 0 ( rejected otherwise z = 0)

Hence,

The required complex number is

$\bf\implies \:z \: = \: \pm \: \dfrac{ \sqrt{3} }{2} - \dfrac{1}{2} i\: or \:z \:= \:i$

Additional Information :-

If z is a complex number, then

$\boxed{ \sf{ \: |z| = |\bar{z}|}}$

$\boxed{ \sf{ \:z \: \bar{z} \: = \: { |z| }^{2}}}$

$\boxed{ \sf{ \:z + \bar{z} = 2 \: Re(z)}}$

$\boxed{ \sf{ \:z - \bar{z} = 2 \: i \: Im(z)}}$

Also,

$\boxed{ \sf{ \: { |z_1 + z_2| }^{2} + { |z_1 - z_2| }^{2} = 2( { |z_1| }^{2} + { |z_2| }^{2})}}$