9.3 Polar form
Product in polar form
\(z_1=r_1(cos(\theta_1)+isin(\theta_1))\)
\(z_2=r_2(cos(\theta_2)+isin(\theta_2))\)
Then: \(z_1\cdot z_2=r_1r_2(cos(\theta_1)cos(\theta_2)-sin(\theta_1)sin(\theta_2)+i(sin(\theta_1)cos(\theta_2)+sin(\theta_2)cos(\theta_1)))\)
\(z_1\cdot z_2=r_1r_2(cos(\theta_1+\theta_2)+isin(\theta_1+\theta_2))\)
In particular, \(z=r(cos(\theta)+isin(\theta))\Rightarrow z^2=r^2(cos(2\theta)+isin(2\theta))\)
Example
\(z=\frac{\sqrt3}{2}+\frac{1}{2}i\)
Compute \(z^3=(\frac{\sqrt3}{2}+\frac{1}{2}i)^3=?\)
Using polar form: \(z=r(cos(\theta)+isin(\theta))\)
\(r=|z|=\sqrt{(3/4+1/4)}=1\)
\(\theta=\frac{\pi}{6}+2k\pi\)
Then, \(z=cos(\frac{\pi}{6})+isin(\frac{\pi}{6})\) and \(z^3=cos(3\frac{\pi}{6})+isin(3\frac{\pi}{6})=cos(\frac{\pi}{2})+isin(\frac{\pi}{2})\)
Describing regions on the plane C
We say use \(Im(z)\) and \(Re(z)\) to describe of C
\(A=\{z\in C:Re(z)=a,a\in R\}\)
\(B=\{z\in C:Im(z)=b,b\in R\}\)
\(C=\{z\in C:Re(z)\geq0,Im(z)\geq0,Im(z)+Re(z)\leq 3,a\in R\}\)
\(z=Re(z)+i\cdot Im(z)\) we identify with \(R^2\)
\(x+y\leq 3\)
Circles and circular sectors
\(D=\{z\in C:|z-z_0|=R,z_0\in C,R>0\}\)
\(E=\{z\in C:r<|z-z_0|<R,z_0\in C,R>0,r>0\}\) \(\frac{\pi}{2}\leq Arg(z)\leq \frac{3\pi}{4}\)
Solving equation in C
Solve \(z^5=-32\)
First, note that \(-32=32(cos(\pi)+isin(\pi))\)
Now,if we write z in polar form: \(z=r(cos(\theta)+isin(\theta))\)
Then, z is a solution if \(r^5=32\Rightarrow r=2\), \(5\theta=\pi+2\pi k ~~k\in Z\)
\(\theta=\frac{\pi+2k\pi}{5}\),\(k\in Z\)
The solutions are \(z_k=2(cos(\theta_k)+isin(\theta_k))\)
If we write \(k=5m+j\) with \(j=0,1,2,3,4\)
\(\theta_k=\frac{\pi+2k\pi}{5}=\frac{\pi+2\pi(5m+j)}{5}=\frac{\pi+2\pi j}{5}+2\pi m\)
\(cos(\theta_k)=cos(\frac{\pi+2j\pi}{5}+2\pi m)=cos(\frac{\pi}{5}+\frac{2\pi j}{5})\)
The solution are \(z_0=2(cos(\frac{\pi}{5})+sin(\frac{\pi}{5}))\),\(z_1=2(cos(\frac{3\pi}{5})+sin(\frac{\pi}{5}))\),\(z_2=-2\),\(z_3=2(cos(\frac{7\pi}{5})+sin(\frac{7\pi}{5})),z_4=2(cos(\frac{9\pi}{5})+sin(\frac{9\pi}{5}))\)
Roots of unity
Consider \(n\in N\). Find the solutions to \(z^n=1=1(cos(0)+isin(0))\)
Writing \(z\) in polar form: \(z=r(cos(\theta)+isin(\theta))\)
Clearly, \(r=1\)
For the angular variable (\(n\in N~fixed\))
\(\theta\in R\) such that \(n\theta=0+2k\pi,k\in Z\)
\(\theta_k=\frac{2\pi k }{n},k\in Z\)
As in the example, we write \(k=n\cdot m+j,j\in\{0,1,2,3,...,n-1\},m\in Z,j\in N\)
\(\theta_k=\frac{2\pi(nm+j)}{n}=\frac{2\pi nm}{n}+\frac{2\pi j}{n}\)
\(cos(\theta_k)=cos(2\pi m+\frac{2\pi j}{n})=cos(\frac{2\pi j}{n})\)
The same with sine: \(sin(\theta_k)=sin(\frac{2\pi j}{n})\)
Solutions: \(z_j=2(cos(\frac{2\pi j}{n})+isin(\frac{2\pi j}{n}))~j=0,1,2,...,n-1\)
Some similar examples
Solve \(z^3=i=1(cos(\frac{\pi}{2})+isin(\frac{\pi}{2}))\)
\(z=1(cos(\theta)+sin(\theta)) with\) \(\theta\in R\)
\(3\theta=\frac{\pi}{2}+2\pi k~~k\in Z\), \(k=3m+j,j=0,1,2\)
Equivalently: \(\theta_j=\frac{\frac{\pi}{2}+2\pi j}{3}, with~j=0,1,2\)
\(\theta_j\in\{\frac{\pi}{6},\frac{5\pi}{6},\frac{3\pi}{2}\}\)
Solutions:
\(z_0=cos(\frac{\pi}{6})+isin(\frac{\pi}{6})=\frac{\sqrt{3}}{2}+\frac{1}{2}i\)
\(z_{1}=\cos(\frac{5}{6}x)+\sin(\frac{5\pi}{6})=-\frac{\sqrt{3}}{2}+\frac{1}{2}i\)
\(z_{2}=\cos(\frac{3\pi}{2})+i\sin(\frac{3\pi}{2})=-i\)
Example
Solve \(z^4=1+i=\sqrt2(cos(\frac{\pi}{4})+isin(\frac{\pi}{4}))\)
Then \(z=r(cos(\theta)+isin(\theta))\) with: \(r^4=2^{1/2}\)
So \(r=2^{\frac{1}{8} }\)
\(\theta \in R\) | \(4\theta=\frac{\pi}{4}+2k\pi,k\in Z\)
\(\theta_j=\frac{\pi+8\pi j}{16}\) with \(j=0,1,2,3\)
Roots of unity, again
\(z^4=1\)
\(z_0=1\)
\(z_1=i\) \(z_1^2=-1\) \(z_1^3=-i\) \(z_1^4=1\)
\(z_2=-1\) \(z_2^2=1\)
\(z_3=-i\) \(z_3^2=-1\) \(z_3^3=i\) \(z_3^4=1\)
\(z^5=1\), \(z_j=cos(\frac{2\pi j}{5})+isin(\frac{2\pi j}{5})\) \(j=0,1,2,3,4\)
for \(j\neq 0\), check that \(z_j^k\neq 1\) \(\forall k=1,2,3,4\)
\(z^6=1\) \(z_j=cos(\frac{2\pi j}{6})+isin(\frac{2\pi j}{6})\) \(j=0,1,2,3,4,5\)
Power of \(z_1\):
\(z_{1}=\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3})\neq1\)
\(z_{1}^{2}=cos(\frac{2\pi}{3})+i\sin(\frac{2\pi}{3})\neq1\)
\(z_{1}^{3}=\cos\left(\pi\right)+i\sin\left(\pi\right)\neq1\)
\(z_{1}^{4}=\cos(\frac{4}{3}\pi)+i\sin(\frac{4\pi}{3})\neq1\)
\(z_{1}^{5}=cos(\frac{5}{3}\pi)+i\sin(\frac{5\pi}{3})\neq1\)
\(z_1^6=1\)
Power of \(z_3\):
\(z_{3}=cos(\pi)+i\sin(\pi)=-1\)
\(z_{3}^{2}=(-1)^{2}=1\)
PRIMITIVE ROOT
We say that a solution \(z\) of the equation \(z^n=1\) is a PRIMITIVE ROOT
If \(z^k\neq1,\forall k=1,2,3,...,n-1\)
Example
All roots from \(z^5=1\) are primitive
If \(n=6\), \(z_1=cos(\frac{\pi}{3})+isin(\frac{\pi}{3})\) is primitive
\(z_3=-1\) is not primitive
Exercise
Find conditions on \(n\in N\) | all solutions of \(z^n=1\) are primitive
Proof
The solution of \(z^n = 1\) is \(z_k = \text{cis} \left( \frac{2k\pi}{n} \right)\), then \(\left(z_{k}\right)^{t}=\text{cis}\left(\frac{2k\pi\cdot t}{n}\right)\) and \(t\) is the power.
By definition of primitive, we need to prove \(z^{t}\neq1,\forall t=1,2,3,...,n-1\)
Which means \(t=min\{n\}\) such that \(z^t=1\)( \(\text{cis}\left(\frac{2k\pi\cdot t}{n}\right)=1\))
Then \(\text{cis}\left(\frac{2k\pi\cdot t}{n}\right)=cis(2m\pi)\), m is an integer
Then \(n \mid kt\) such that \(\frac{kt}{n}\) is an integer, if and only if at least \(t = n\). \(k \in [0, n-1]\), \(t\in[1,n]\)
- If \(n\) is a prime number
By definition of prime, only when \(t = n\), \(n \mid kt\).
Thus \(n\) can be prime.
-
If \(n\) isn't a prime number
We need to prove \(\forall n\in N,n\) is not a prime such that all solutions of \(z^n=1\) are not primitive
Suppose \(\exist n\in N\), \(n\) is not a prime such that all solutions of \(z^n=1\) are primitive
Then by fundamental theorem of arithmetic, we know \(n=p_1p_2\ldots p_{n}\)(\(p_1p_2...p_n\) are all prime)
Due to exist one n that is primitive, then \(n\mid kt\), iff \(t=n\)
Thus \(p_1p_2\ldots p_{n}\mid kt\) and \(p_j\nmid kt\) for all \(1\leq j\leq n\)
But by the property of integer divisibility(\(p_1p_2\ldots p_{n}\) are prime numbers), \(p_j\mid kt\) for all \(1\leq j\leq n-1\) must be true.
Contradiction!
Hence \(n\) must be a prime number.