nth Root of a Complex Number

N-th Root of a Complex Number \( \bf (z)\) #

  Statement: Let \(\bf z=r(\cos\theta +i \sin\theta)\), where \(\bf -\pi~\text{\textless}~\theta~\le~\pi \), is a non-complex number. If \(\bf n\) be a positive integer then there exists \(\bf n\) distinct values of \(\bf z^{\frac{1}{n}}\) .
Proof:

 Given that \(\bf z=r(\cos\theta +i \sin\theta)\), where \(\bf -\pi~\text{\textless}~\theta~\le~\pi \), is a non-complex number.
 Let \(\bf n\) is a positive integer.
 To prove \(\bf z^{\frac{1}{n}}\) has \(\bf n\) distinct values.
 Let \begin{align*} &\bf z^{\frac{1}{n}}=\rho(\cos\phi +i \sin\phi)\\ \implies &\bf z=\rho^{n}(\cos\phi +i \sin\phi)^{n}\\ \end{align*}  using De Moivre’s theorem, we get \begin{align} &\bf r(\cos\theta +i \sin\theta)=\rho^{n}(\cos n\phi +i \sin n\phi)\nonumber\\ \implies &\bf r\cos\theta=\rho^{n}\cos n\phi\nonumber\\ \bf and~&\bf r\sin\theta=\rho^{n}\sin n\phi \tag{1} \end{align}  Squaring and adding, we get \(\bf r^{2}=\rho^{2n} \) i.e., \(\bf r=\rho^{n}\) or \(\bf \rho=r^{\frac{1}{n}}\). Then from (1) \begin{align*} &\bf \cos\theta=\cos n\phi\nonumber\\ \bf and~&\bf \sin\theta=\sin n\phi \nonumber\\ \implies &\bf n\phi=2k\pi+\theta~,~k\in\Z \nonumber\\ \implies &\bf \phi=\frac{2k\pi+\theta}{n}~,~k\in\Z \end{align*}  Let $$\bf z_{k}=r^{\frac{1}{n}}\left[\cos\left(\frac{2k\pi+\theta}{n} \right) +i \sin\left(\frac{2k\pi+\theta}{n} \right) \right],~k\in\Z $$  We claim that \(\bf z_{0},z_{1},…,z_{n-1} \) are all distinct.
 Since if  \(\bf \exists~p,q \) where \( \bf 0~\le p~\text{\textless}~q\le~ n-1 \) such that \begin{align*} &\bf z_{q}=z_{q} \\ \implies &\bf r^{\frac{1}{n}}\left[\cos\left(\frac{2q\pi+\theta}{n} \right) +i \sin\left(\frac{2q\pi+\theta}{n} \right)\right]\\ &\bf=r^{\frac{1}{n}}\left[\cos\left(\frac{2p\pi+\theta}{n} \right) +i \sin\left(\frac{2p\pi+\theta}{n} \right) \right]\\ \implies&\bf \cos\left(\frac{2q\pi+\theta}{n} \right)=\cos\left(\frac{2p\pi+\theta}{n} \right)\\ \bf and~&\bf \sin\left(\frac{2q\pi+\theta}{n} \right)=\sin\left(\frac{2p\pi+\theta}{n} \right)\\ \implies&\bf \frac{2q\pi+\theta}{n}=\frac{2p\pi+\theta}{n}+2s\pi~,~s\in \Z\\ \implies&\bf \frac{2q\pi+\theta}{n}-\frac{2p\pi+\theta}{n}=2s\pi~,~s\in \Z\\ \implies&\bf \frac{q-p}{n}=s~,~s\in \Z \end{align*}  A contradiction since the L.H.S is a pure fraction but the R.H.S is an integer. Therefore \(\bf z_{0},z_{1},…,z_{n-1} \) are all distinct.
 Let \(\bf m\in \Z \) be any integer. Then by division algorithm for \(\bf m\) and \(\bf n>0\) \(\bf \exists\) two integers \(\bf v\) and \(\bf d\) such that \(\bf m=nv+d\) where \(\bf 0~\le~ d~ \le~ n-1\). Therefore \begin{align*} \bf z_{m} & \bf=r^{\frac{1}{n}} \left[\cos\left(\frac{2m\pi+\theta}{n} \right) +i \sin\left(\frac{2m\pi+\theta}{n} \right)\right] \\ & \bf=r^{\frac{1}{n}} \left[\cos\left(\frac{2nv\pi+2r\pi+\theta}{n} \right) +i \sin\left(\frac{2nv\pi+2r\pi+\theta}{n} \right)\right] \\ & \bf=r^{\frac{1}{n}}\left[\cos\left(2v\pi+\frac{2r\pi+\theta}{n} \right) +i \sin\left(2v\pi+\frac{2r\pi+\theta}{n} \right)\right] \\ & \bf=r^{\frac{1}{n}} \left[\cos\left(\frac{2r\pi+\theta}{n} \right) +i \sin\left(\frac{2r\pi+\theta}{n} \right)\right] \\ & \bf=z_{d}~,~0~\le~ d~ \le~ n-1 \end{align*}  Hence \(\bf z^{\frac{1}{n}}\) has \(\bf n\) distinct values and the values are $$\bf r^{\frac{1}{n}}\left[\cos\left(\frac{2k\pi+\theta}{n} \right) +i \sin\left(\frac{2k\pi+\theta}{n} \right) \right] $$ where \(\bf k=0,1,2,…,n-1 \).

N-th Root of a Complex Number \( \bf (z^{m})\) #

 Statement: Let \(\bf z=r(\cos\theta +i \sin\theta)\), where \(\bf -\pi~\text{\textless}~\theta~\le~\pi \), is a non-complex number. If \(\bf n\) be a positive integer then there exists \(\bf n\) distinct values of \(\bf z^{\frac{m}{n}}\) where \( \bf gcd(m,n)=1\).
Proof: 

 Given that \(\bf z=r(\cos\theta +i \sin\theta)\), where \(\bf -\pi~\text{\textless}~\theta~\le~\pi \), is a non-complex number.
 Let \(\bf n\) is a positive integer.
 To prove \(\bf z^{\frac{m}{n}}\) has \(\bf n\) distinct values where \( \bf gcd(m,n)=1\).
 Let \begin{align*} &\bf z^{\frac{m}{n}}=\rho(\cos\phi +i \sin\phi)\\ \implies &\bf z^{m}=\rho^{n}(\cos\phi +i \sin\phi)^{n}\\ \end{align*}  using De Moivre’s theorem, we get \begin{align} &\bf r^{m}(\cos m\theta +i \sin m\theta)=\rho^{n}(\cos n\phi +i \sin n\phi)\nonumber\\ \implies &\bf r^{m}\cos m\theta=\rho^{n}\cos n\phi\nonumber\\ \bf and~&\bf r^{m}\sin m\theta=\rho^{n}\sin n\phi \tag{1} \end{align}  Squaring and adding, we get \(\bf r^{2m}=\rho^{2n} \) i.e., \(\bf r^{m}=\rho^{n}\) or \(\bf \rho=r^{\frac{m}{n}}\). Then from (1) \begin{align*} &\bf \cos m\theta=\cos n\phi\nonumber\\ \bf and~&\bf \sin m\theta=\sin n\phi \nonumber\\ \implies &\bf n\phi=2k\pi+ m\theta~,~k\in\Z \nonumber\\ \implies &\bf \phi=\frac{2k\pi+ m\theta}{n}~,~k\in\Z \end{align*}  Let $$\bf z_{k}=r^{\frac{1}{n}}\left[\cos\left(\frac{2k\pi+ m\theta}{n} \right) +i \sin\left(\frac{2k\pi+ m\theta}{n} \right) \right],~k\in\Z $$  We claim that \(\bf z_{0},z_{1},…,z_{n-1} \) are all distinct.
 Since if  \(\bf \exists~p,q \) where \( \bf 0~\le p~\text{\textless}~ q~\le~ n-1 \) such that \begin{align*} &\bf z_{q}=z_{q} \\ \implies &\bf r^{\frac{m}{n}}\left[\cos\left(\frac{2q\pi+ m\theta}{n} \right) +i \sin\left(\frac{2q\pi+ m\theta}{n} \right)\right]\\ &\bf=r^{\frac{m}{n}}\left[\cos\left(\frac{2p\pi+ m\theta}{n} \right) +i \sin\left(\frac{2p\pi+ m\theta}{n} \right) \right]\\ \implies&\bf \cos\left(\frac{2q\pi+ m\theta}{n} \right)=\cos\left(\frac{2p\pi+ m\theta}{n} \right)\\ \bf and~&\bf \sin\left(\frac{2q\pi+ m\theta}{n} \right)=\sin\left(\frac{2p\pi+ m\theta}{n} \right)\\ \implies&\bf \frac{2q\pi+ m\theta}{n}=\frac{2p\pi+ m\theta}{n}+2s\pi~,~s\in \Z\\ \implies&\bf \frac{2q\pi+ m\theta}{n}-\frac{2p\pi+ m\theta}{n}=2s\pi~,~s\in \Z\\ \implies&\bf \frac{q-p}{n}=s~,~s\in \Z \end{align*}  A contradiction since the L.H.S is a pure fraction but the R.H.S is an integer. Therefore \(\bf z_{0},z_{1},…,z_{n-1} \) are all distinct.
 Let \(\bf t\in \Z \) be any integer. Then by division algorithm for \(\bf m\) and \(\bf n>0\) \(\bf \exists\) two integers \(\bf v\) and \(\bf r\) such that \(\bf t=nv+d\) where \(\bf 0~\le~ d~ \le~ n-1\). Therefore \begin{align*} \bf z_{t} & \bf=r^{\frac{m}{n}} \left[\cos\left(\frac{2t\pi+ m\theta}{n} \right) +i \sin\left(\frac{2t\pi+ m\theta}{n} \right)\right] \\ & \bf=r^{\frac{m}{n}} \left[\cos\left(\frac{2nv\pi+2r\pi+ m\theta}{n} \right) +i \sin\left(\frac{2nv\pi+2r\pi+ m\theta}{n} \right)\right] \\ & \bf=r^{\frac{m}{n}}\left[\cos\left(2v\pi+\frac{2r\pi+ m\theta}{n} \right) +i \sin\left(2v\pi+\frac{2r\pi+ m\theta}{n} \right)\right] \\ & \bf=r^{\frac{m}{n}} \left[\cos\left(\frac{2r\pi+ m\theta}{n} \right) +i \sin\left(\frac{2r\pi+ m\theta}{n} \right)\right] \\ & \bf=z_{d}~,~0~\le~ d~ \le~ n-1 \end{align*}  Hence \(\bf z^{\frac{m}{n}}\) has \(\bf n\) distinct values and the values are $$\bf r^{\frac{m}{n}}\left[\cos\left(\frac{2k\pi+ m\theta}{n} \right) +i \sin\left(\frac{2k\pi+ m\theta}{n} \right) \right] $$ where \(\bf k=0,1,2,…,n-1 \).

N-th Root of Unity \( \bf (z=1)\) #

 Statement: Let \(\bf z=1\). If \(\bf n\) be a positive integer then there exists \(\bf n\) distinct values of \(\bf 1^{\frac{1}{n}}\) .
Proof: 

 Given that \(\bf z=1\).
 Let \(\bf n\) is a positive integer.
 To prove \(\bf 1^{\frac{1}{n}}\) has \(\bf n\) distinct values.
 Let \begin{align*} &\bf 1^{\frac{1}{n}}=\rho(\cos\phi +i \sin\phi)\\ \implies &\bf 1=\rho^{n}(\cos\phi +i \sin\phi)^{n}\\ \end{align*}  using De Moivre’s theorem, we get \begin{align} &\bf 1=\rho^{n}(\cos n\phi +i \sin n\phi)\nonumber\\ \implies &\bf 1=\rho^{n}\cos n\phi\nonumber\\ \bf and~&\bf 0=\rho^{n}\sin n\phi \tag{1} \end{align}  Squaring and adding, we get \(\bf 1=\rho^{2n} \) i.e., \(\bf 1=\rho^{n}\) or \(\bf \rho=1\). Then from (1) \begin{align*} &\bf1=\cos n\phi\nonumber\\ \bf and~&\bf 0=\sin n\phi \nonumber\\ \implies &\bf n\phi=2k\pi~,~k\in\Z \nonumber\\ \implies &\bf \phi=\frac{2k\pi}{n}~,~k\in\Z \end{align*}  Let $$\bf z_{k}=\left[\cos\left(\frac{2k\pi}{n} \right) +i \sin\left(\frac{2k\pi}{n} \right) \right],~k\in\Z $$  We claim that \(\bf z_{0},z_{1},…,z_{n-1} \) are all distinct.
 Since if  \(\bf \exists~p,q \) where \( \bf 0~\le p~\text{\textless}~q\le~ n-1 \) such that \begin{align*} &\bf z_{q}=z_{q} \\ \implies &\bf \left[\cos\left(\frac{2q\pi}{n} \right) +i \sin\left(\frac{2q\pi}{n} \right)\right]\\ &\bf=\left[\cos\left(\frac{2p\pi}{n} \right) +i \sin\left(\frac{2p\pi}{n} \right) \right]\\ \implies&\bf \cos\left(\frac{2q\pi}{n} \right)=\cos\left(\frac{2p\pi}{n} \right)\\ \bf and~&\bf \sin\left(\frac{2q\pi}{n} \right)=\sin\left(\frac{2p\pi}{n} \right)\\ \implies&\bf \frac{2q\pi}{n}=\frac{2p\pi}{n}+2s\pi~,~s\in \Z\\ \implies&\bf \frac{2q\pi}{n}-\frac{2p\pi}{n}=2s\pi~,~s\in \Z\\ \implies&\bf \frac{q-p}{n}=s~,~s\in \Z \end{align*}  A contradiction since the L.H.S is a pure fraction but the R.H.S is an integer. Therefore \(\bf z_{0},z_{1},…,z_{n-1} \) are all distinct.
 Let \(\bf m\in \Z \) be any integer. Then by division algorithm for \(\bf m\) and \(\bf n>0\) \(\bf \exists\) two integers \(\bf v\) and \(\bf r\) such that \(\bf m=nv+r\) where \(\bf 0~\le~ r~ \le~ n-1\). Therefore \begin{align*} \bf z_{m} & \bf=\left[\cos\left(\frac{2m\pi}{n} \right) +i \sin\left(\frac{2m\pi}{n} \right)\right] \\ & \bf= \left[\cos\left(\frac{2nv\pi+2r\pi}{n} \right) +i \sin\left(\frac{2nv\pi+2r\pi}{n} \right)\right] \\ & \bf=\left[\cos\left(2v\pi+\frac{2r\pi}{n} \right) +i \sin\left(2v\pi+\frac{2r\pi}{n} \right)\right] \\ & \bf= \left[\cos\left(\frac{2r\pi}{n} \right) +i \sin\left(\frac{2r\pi}{n} \right)\right] \\ & \bf=z_{r}~,~0~\le~ r~ \le~ n-1 \end{align*}  Hence \(\bf z^{\frac{1}{n}}\) has \(\bf n\) distinct values and the values are $$\bf \left[\cos\left(\frac{2k\pi}{n} \right) +i \sin\left(\frac{2k\pi}{n} \right) \right] $$ where \(\bf k=0,1,2,…,n-1 \).