Multiplying and dividing complex numbers
Method
- If \(z_{1}=r_{1}e^{i\theta_{1}}\) and \(z_{2}=r_{2}e^{i\theta_{2}}\) then:
- \(z_{1}z_{2}=r_{1}r_{2}e^{i(\theta_{1}+\theta_{2})}\)
- \(\frac{z_{1}}{z_{2}}=\frac{r_{1}}{r_{2}}e^{i(\theta_{1}-\theta_{2})}\)
Example
\(3\left( \cos\frac{5\pi}{12}+i\sin\frac{5\pi}{12} \right)\times4\left( \cos \frac{\pi}{12}+i\sin \frac{\pi}{12} \right)\)
\(=12\left( \cos \frac{\pi}{2}+i\sin \frac{\pi}{2} \right)\)
\(=12(0+(i\times1))\)
\(=12i\)
Find \(z^5\)
- Modulus of 2
- Argument of \(\frac{1}{3}\pi\)
- Written as \(2\left( \cos \frac{1}{6}\pi+\sin \frac{1}{6}\pi \right)\)
- Or as \((2e^{i \frac{\pi}{6}})^5\)
- Which equals \(32 e^{i \frac{5\pi}{6}}\)
\(z^n=r^n(\cos n\theta+i\sin n\theta)\)
- Show true for \(n=1\)
- \(z^1=r^1(\cos\theta+i\sin\theta)\)
- Assume true for \(n=k\)
- \(z^k=r^k(\cos k\theta+i\sin k\theta)\)
- Show true for \(n=k+1\)
- \(z^{k+1}=r^{k+1}(\cos (k+1)\theta+i\sin (k+1)\theta)\)
- \(z^{k+1}=z^k+z\)
- \(=r^k(\cos k\theta+i\sin k\theta)\times r(\cos\theta+i\sin\theta)\)
- \(=r^{k+1}(\cos (k\theta+\theta)+i\sin(k\theta+\theta))\)
- \(=r^{k+1}(\cos ((k+1)\theta)+i\sin((k+1)\theta))\)
- Hence true for all positive n.