Question

How to solve the equation \[{{z}^{4}}=i\overline{z}\] , where z is a complex number?

Answer 1

Hint: To solve the equation \[{{z}^{4}}=i\overline{z}\] , we have to assume $z=r{{e}^{i\theta }}$ . Then, $\overline{z}$ is the complex conjugate, that is, $\overline{z}=r{{e}^{-i\theta }}$ . Now, we have to substitute these values in the given equation as simplify. We will change the exponent because complex exponents have a period of $2\pi $ . So, we will get ${{\left( r{{e}^{i\theta }} \right)}^{4}}=i\left( r{{e}^{i\left( -\theta +2n\pi \right)}} \right),n\in \mathbb{N}$ . Now, we will further simplify this equation and equate the modulus and argument parts (or real and imaginary parts). Then, we will find the value of r and corresponding values of z for various values of n.

Complete step-by-step solution:
We have to solve the equation \[{{z}^{4}}=i\overline{z}\] . We are given that z is a complex number. Let us assume $z=r{{e}^{i\theta }}$ . Then, $\overline{z}$ is the complex conjugate, that is, $\overline{z}=r{{e}^{-i\theta }}$ . Now, let us substitute these in the given equation.
$\Rightarrow {{\left( r{{e}^{i\theta }} \right)}^{4}}=i\left( r{{e}^{-i\theta }} \right)$
We know that complex exponents have has a period of $2\pi $ . Therefore, we can write the above equation as
$\Rightarrow {{\left( r{{e}^{i\theta }} \right)}^{4}}=i\left( r{{e}^{i\left( -\theta +2n\pi \right)}} \right),n\in \mathbb{N}$
We know that ${{e}^{i\dfrac{\pi }{2}}}=\cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2}=i$ .
$\Rightarrow {{\left( r{{e}^{i\theta }} \right)}^{4}}={{e}^{i\dfrac{\pi }{2}}}\left( r{{e}^{i\left( -\theta +2n\pi \right)}} \right)$
We know that ${{\left( ab \right)}^{m}}={{a}^{m}}\times {{b}^{m}}$ and ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ .
$\Rightarrow {{r}^{4}}{{e}^{i4\theta }}={{e}^{i\dfrac{\pi }{2}}}\left( r{{e}^{i\left( -\theta +2n\pi \right)}} \right)$
We know that ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ .
$\Rightarrow {{r}^{4}}{{e}^{i4\theta }}=r{{e}^{i\left( \dfrac{\pi }{2}-\theta +2n\pi \right)}}$
Now, let us equate the modulus and argument parts (or real and imaginary parts). When we equate the real parts, we will get
${{r}^{4}}=r...\left( i \right)$
When we equate the imaginary, we will get
$4\theta =\dfrac{\pi }{2}-\theta +2n\pi ...\left( ii \right)$
From equation (i), we can find the value of r. Let us take r from the RHS to the LHS.
$\Rightarrow {{r}^{4}}-r=0$
Let us take common r outside.
$\Rightarrow r\left( {{r}^{3}}-1 \right)=0$
This means, we can write either $r=0$ or ${{r}^{3}}-1=0$ . Let us consider ${{r}^{3}}-1=0$ . We have to take -1 to the RHS.
$\Rightarrow {{r}^{3}}=1$
Let us take the cube root of the above equation.
$\Rightarrow r=1$
Therefore, we obtained $r=0,1$ .
So, when $r=0$ , we can write
$\begin{align}
  & z=r{{e}^{i\theta }}=0\times {{e}^{i\theta }} \\
 & \Rightarrow z=0 \\
\end{align}$
When $r=1$ , we will get
$\begin{align}
  & z=r{{e}^{i\theta }}=1\times {{e}^{i\theta }} \\
 & \Rightarrow z={{e}^{i\theta }} \\
\end{align}$
Now, let us simplify the equation (ii) by taking $\theta $ to the LHS.
$\begin{align}
  & \Rightarrow 4\theta +\theta =\dfrac{\pi }{2}+2n\pi \\
 & \Rightarrow 5\theta =\dfrac{\pi }{2}+2n\pi \\
\end{align}$
Let us simplify the RHS.
$\begin{align}
  & \Rightarrow 5\theta =\dfrac{\pi }{2}+\dfrac{4n\pi }{2} \\
 & \Rightarrow 5\theta =\dfrac{\pi }{2}\left( 1+4n \right) \\
\end{align}$
We have to take 5 to the RHS.
$\Rightarrow \theta =\dfrac{\pi }{10}\left( 1+4n \right)$
Let us substitute various values of n, that is, 0, 1, 2, … in the above equation and find the corresponding values of z when $r=1$, that is, $z={{e}^{i\theta }}$ .
We know that ${{e}^{i\theta }}=\cos \theta +i\sin \theta $ .
$\Rightarrow z=\cos \theta +i\sin \theta $
When $n=0$ , we will get
$\theta =\dfrac{\pi }{10}$
$\therefore z=\cos \dfrac{\pi }{10}+i\sin \dfrac{\pi }{10}$
We know that
 $\begin{align}
  & \Rightarrow \cos x=2{{\cos }^{2}}\dfrac{x}{2}-1 \\
 & \Rightarrow {{\cos }^{2}}\dfrac{x}{2}=\dfrac{\cos x+1}{2} \\
 & \Rightarrow \cos \dfrac{x}{2}=\sqrt{\dfrac{\cos x+1}{2}} \\
\end{align}$
Now, we can write $\cos \dfrac{\pi }{10}=\cos \dfrac{\left( \dfrac{\pi }{5} \right)}{2}=\sqrt{\dfrac{\cos \dfrac{\pi }{5}+1}{2}}...\left( iii \right)$

We know that
 $\begin{align}
  & \dfrac{2\pi }{5}=\pi -\dfrac{3\pi }{5} \\
 & \Rightarrow \sin \left( \dfrac{2\pi }{5} \right)=\sin \left( \pi -\dfrac{3\pi }{5} \right) \\
 & \Rightarrow \sin \left( \dfrac{2\pi }{5} \right)=\sin \left( \dfrac{3\pi }{5} \right) \\
\end{align}$
We know that $\sin 2\theta =2\sin \theta \cos \theta $ .
$\begin{align}
  & \Rightarrow 2\sin \left( \dfrac{\pi }{5} \right)\cos \left( \dfrac{\pi }{5} \right)=3\sin \left( \dfrac{\pi }{5} \right)-4{{\sin }^{3}}\left( \dfrac{\pi }{5} \right) \\
 & \Rightarrow 2\sin \left( \dfrac{\pi }{5} \right)\cos \left( \dfrac{\pi }{5} \right)=\sin \left( \dfrac{\pi }{5} \right)\left( 3-4{{\sin }^{2}}\left( \dfrac{\pi }{5} \right) \right) \\
 & \Rightarrow 2\require{cancel}\cancel{\sin \left( \dfrac{\pi }{5} \right)}\cos \left( \dfrac{\pi }{5} \right)=\require{cancel}\cancel{\sin \left( \dfrac{\pi }{5} \right)}\left( 3-4{{\sin }^{2}}\left( \dfrac{\pi }{5} \right) \right) \\
 & \Rightarrow 2\cos \left( \dfrac{\pi }{5} \right)=3-4{{\sin }^{2}}\left( \dfrac{\pi }{5} \right) \\
\end{align}$
We know that ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ .
$\Rightarrow 2\cos \left( \dfrac{\pi }{5} \right)=3-4\left( 1-{{\cos }^{2}}\left( \dfrac{\pi }{5} \right) \right)$
Let us consider $X=\cos \left( \dfrac{\pi }{5} \right)$ .
$\begin{align}
  & \Rightarrow 2X=3-4\left( 1-{{X}^{2}} \right) \\
 & \Rightarrow 2X=3-4+4{{X}^{2}} \\
 & \Rightarrow 2X=-1+4{{X}^{2}} \\
 & \Rightarrow 4{{X}^{2}}-2X-1 \\
 & \Rightarrow 4{{X}^{2}}-2X-1 \\
\end{align}$
Let us substitute $4{{X}^{2}}-2X+1=0$ . Then using quadratic formula, we can find the value of X.
$\begin{align}
  & X=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\times 4\times -1}}{2\times 4} \\
 & \Rightarrow X=\dfrac{2\pm \sqrt{4+16}}{8} \\
 & \Rightarrow X=\dfrac{2\pm \sqrt{20}}{8} \\
 & \Rightarrow X=\dfrac{2\pm 2\sqrt{5}}{8} \\
 & \Rightarrow X=\dfrac{1\pm \sqrt{5}}{4} \\
 & \Rightarrow X=\dfrac{1+\sqrt{5}}{4},X=\dfrac{1-\sqrt{5}}{4} \\
\end{align}$
We know that $\cos \left( \dfrac{\pi }{5} \right)$ is positive.
$\begin{align}
  & \Rightarrow X=\dfrac{1+\sqrt{5}}{4} \\
 & \Rightarrow \cos \left( \dfrac{\pi }{5} \right)=\dfrac{1+\sqrt{5}}{4} \\
\end{align}$
Let us substitute this in (iii).
$\begin{align}
  & \cos \dfrac{\pi }{10}==\sqrt{\dfrac{\dfrac{1+\sqrt{5}}{4}+1}{2}} \\
 & \Rightarrow \cos \dfrac{\pi }{10}=\sqrt{\dfrac{1+\sqrt{5}+4}{8}} \\
 & \Rightarrow \cos \dfrac{\pi }{10}=\sqrt{\dfrac{5+\sqrt{5}}{8}} \\
 & \Rightarrow \cos \dfrac{\pi }{10}=\dfrac{\sqrt{5+\sqrt{5}}}{2\sqrt{2}} \\
 & \Rightarrow \cos \dfrac{\pi }{10}=\dfrac{1}{2}\sqrt{\dfrac{1}{2}\left( 5+\sqrt{5} \right)} \\
\end{align}$
Now, let us find $\sin \dfrac{\pi }{10}$ .
$\sin \dfrac{\pi }{10}=\sin \dfrac{\left( \dfrac{\pi }{5} \right)}{2}$
We know that
 $\begin{align}
  & \sin 2x=2\sin x\cos x \\
 & \Rightarrow \sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2} \\
 & \Rightarrow \sin \dfrac{x}{2}=\dfrac{\sin x}{2\cos \dfrac{x}{2}} \\
\end{align}$
Therefore, we can write
$\sin \dfrac{\left( \dfrac{\pi }{5} \right)}{2}=\dfrac{\sin \dfrac{\pi }{5}}{2\cos \dfrac{\pi }{10}}$
We obtained $\cos \left( \dfrac{\pi }{5} \right)=\dfrac{1+\sqrt{5}}{4}$ .
We know that ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ .
$\begin{align}
  & \Rightarrow {{\sin }^{2}}\left( \dfrac{\pi }{5} \right)=1-{{\left( \dfrac{1+\sqrt{5}}{4} \right)}^{2}} \\
 & \Rightarrow {{\sin }^{2}}\left( \dfrac{\pi }{5} \right)=1-\left( \dfrac{1-2\sqrt{5}+{{\left( \sqrt{5} \right)}^{2}}}{16} \right) \\
 & \Rightarrow {{\sin }^{2}}\left( \dfrac{\pi }{5} \right)=1-\left( \dfrac{6-2\sqrt{5}}{16} \right) \\
 & \Rightarrow {{\sin }^{2}}\left( \dfrac{\pi }{5} \right)=1-\left( \dfrac{3-\sqrt{5}}{8} \right) \\
 & \Rightarrow {{\sin }^{2}}\left( \dfrac{\pi }{5} \right)=\left( \dfrac{8-3+\sqrt{5}}{8} \right) \\
 & \Rightarrow {{\sin }^{2}}\left( \dfrac{\pi }{5} \right)=\dfrac{5-\sqrt{5}}{8} \\
 & \Rightarrow \sin \left( \dfrac{\pi }{5} \right)=\sqrt{\dfrac{5-\sqrt{5}}{8}} \\
\end{align}$
Therefore, we can write
$\begin{align}
  & \Rightarrow \sin \dfrac{\left( \dfrac{\pi }{5} \right)}{2}=\dfrac{\sqrt{\dfrac{5-\sqrt{5}}{8}}}{2\cos \dfrac{\pi }{10}} \\
 & \Rightarrow \sin \dfrac{\left( \dfrac{\pi }{5} \right)}{2}=\dfrac{\sqrt{\dfrac{5-\sqrt{5}}{8}}}{2\sqrt{\dfrac{5+\sqrt{5}}{8}}} \\
 & \Rightarrow \sin \dfrac{\pi }{10}=\dfrac{\sqrt{\dfrac{5-\sqrt{5}}{8}}}{2\times \dfrac{1}{2}\sqrt{\dfrac{1}{2}\left( 5+\sqrt{5} \right)}} \\
 & \Rightarrow \sin \dfrac{\pi }{10}=\sqrt{\dfrac{5-\sqrt{5}}{4\left( 5+\sqrt{5} \right)}} \\
\end{align}$
Let us take the conjugate.
$\begin{align}
  & \Rightarrow \sin \dfrac{\pi }{10}=\dfrac{1}{2}\sqrt{\dfrac{\left( 5-\sqrt{5} \right)\left( 5-\sqrt{5} \right)}{\left( 5+\sqrt{5} \right)\left( 5-\sqrt{5} \right)}} \\
 & \Rightarrow \sin \dfrac{\pi }{10}=\dfrac{1}{2}\sqrt{\dfrac{{{\left( 5-\sqrt{5} \right)}^{2}}}{\left( 5+\sqrt{5} \right)\left( 5-\sqrt{5} \right)}} \\
\end{align}$
$\Rightarrow \sin \dfrac{\pi }{10}=\dfrac{1}{2}\sqrt{\dfrac{\left( 5-\sqrt{5} \right)\left( 5-\sqrt{5} \right)}{\left( 5+\sqrt{5} \right)\left( 5-\sqrt{5} \right)}}$
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ and $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$ .
$\begin{align}
  & \Rightarrow \sin \dfrac{\pi }{10}=\dfrac{1}{2}\sqrt{\dfrac{{{\left( 5-\sqrt{5} \right)}^{2}}}{\left( 5+\sqrt{5} \right)\left( 5-\sqrt{5} \right)}} \\
 & \Rightarrow \sin \dfrac{\pi }{10}=\dfrac{1}{2}\sqrt{\dfrac{25-10\sqrt{5}+5}{25-5}} \\
 & \Rightarrow \sin \dfrac{\pi }{10}=\dfrac{1}{2}\sqrt{\dfrac{30-10\sqrt{5}}{20}} \\
 & \Rightarrow \sin \dfrac{\pi }{10}=\dfrac{1}{2}\sqrt{\dfrac{3-\sqrt{5}}{2}} \\
\end{align}$
Now, we can write z as
$z=\dfrac{1}{2}\sqrt{\dfrac{1}{2}\left( 5+\sqrt{5} \right)}+i\dfrac{1}{2}\sqrt{\dfrac{3-\sqrt{5}}{2}}$

When $n=1$ ,
$\begin{align}
  & \Rightarrow \theta =\dfrac{\pi }{10}\left( 1+4 \right) \\
 & \Rightarrow \theta =\dfrac{\pi }{10}\times 5 \\
 & \Rightarrow \theta =\dfrac{\pi }{2} \\
\end{align}$
$\begin{align}
  & \therefore z=\cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2} \\
 & \Rightarrow z=1+0 \\
 & \Rightarrow z=1 \\
\end{align}$
When $n=2$ ,
$\begin{align}
  & \Rightarrow \theta =\dfrac{\pi }{10}\left( 1+4\times 2 \right) \\
 & \Rightarrow \theta =\dfrac{\pi }{10}\times \left( 1+8 \right) \\
 & \Rightarrow \theta =\dfrac{9\pi }{10} \\
\end{align}$
We will get z as
$z=\cos \dfrac{9\pi }{10}+i\sin \dfrac{9\pi }{10}$
We have to evaluate similar to the above steps. We will get
$z=-\dfrac{1}{2}\sqrt{\dfrac{1}{2}\left( 5+\sqrt{5} \right)}+i\dfrac{1}{2}\sqrt{\dfrac{3-\sqrt{5}}{2}}$
When $n=3$ ,
$\begin{align}
  & \Rightarrow \theta =\dfrac{\pi }{10}\left( 1+4\times 3 \right) \\
 & \Rightarrow \theta =\dfrac{\pi }{10}\times \left( 1+12 \right) \\
 & \Rightarrow \theta =\dfrac{13\pi }{10} \\
\end{align}$
We will get z as
$z=\cos \dfrac{13\pi }{10}+i\sin \dfrac{13\pi }{10}$
We have to evaluate similar to the above steps. We will get
$z=-\dfrac{1}{2}\sqrt{\dfrac{1}{2}\left( 5+\sqrt{5} \right)}-i\dfrac{1}{2}\sqrt{\dfrac{3-\sqrt{5}}{2}}$
When $n=4$ ,
$\begin{align}
  & \Rightarrow \theta =\dfrac{\pi }{10}\left( 1+4\times 4 \right) \\
 & \Rightarrow \theta =\dfrac{\pi }{10}\times \left( 1+16 \right) \\
 & \Rightarrow \theta =\dfrac{17\pi }{10} \\
\end{align}$
We will get z as
$z=\cos \dfrac{17\pi }{10}+i\sin \dfrac{17\pi }{10}$
We have to evaluate similar to the above steps. We will get
$z=\dfrac{1}{2}\sqrt{\dfrac{1}{2}\left( 5+\sqrt{5} \right)}-i\dfrac{1}{2}\sqrt{\dfrac{3-\sqrt{5}}{2}}$
This process repeats. Therefore, we will get six value of z as
$z=0,i,\pm \dfrac{1}{2}\sqrt{\dfrac{1}{2}\left( 5+\sqrt{5} \right)}+i\dfrac{1}{2}\sqrt{\dfrac{3-\sqrt{5}}{2}},z=\pm \dfrac{1}{2}\sqrt{\dfrac{1}{2}\left( 5+\sqrt{5} \right)}-i\dfrac{1}{2}\sqrt{\dfrac{3-\sqrt{5}}{2}}$

Note: Students must know to simplify the exponents. They must know to find the values of trigonometric functions using the identities. Students must know that the complex number,z can be written as $z=r{{e}^{i\theta }}$ or $z=x+iy$.