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Mathematics · Ch 10 — Complex Numbers

Exponential Form

10.5.5

Exponential Form

Exponential Form

It is known (and can be proved using special infinite series — a deeper result not derived here) that

eiθ=cos⁡θ+isin⁡θe^{i\theta}=\cos\theta+i\sin\theta

This is Euler's identity. Combining it with the polar form (Section 1.5.4):

z=a+ib=r(cos⁡θ+isin⁡θ)=reiθz=a+ib=r(\cos\theta+i\sin\theta)=re^{i\theta}

where r=∣z∣r=|z| and θ=arg⁡z\theta=\arg z, this is called the exponential form of the complex number.

Continuing the previous worked example, the same four numbers converted to exponential form:

  1. z=4+43i=8eiπ/3z=4+4\sqrt3i=8e^{i\pi/3}.
  2. z=−2=2eiπz=-2=2e^{i\pi}.
  3. z=3i=3eiπ/2z=3i=3e^{i\pi/2}.
  4. z=−3+i=2e5πi/6z=-\sqrt3+i=2e^{5\pi i/6}. Worked Example: express z=2 e3πi/4z=\sqrt2\,e^{3\pi i/4} in a+iba+ib form. Here r=2, θ=3π4r=\sqrt2,\ \theta=\dfrac{3\pi}{4}. The polar form is z=2(cos⁡3π4+isin⁡3π4)z=\sqrt2\left(\cos\dfrac{3\pi}{4}+i\sin\dfrac{3\pi}{4}\right). Using allied angles: cos⁡3π4=cos⁡(π−π4)=−cos⁡π4=−12\cos\dfrac{3\pi}{4}=\cos\left(\pi-\dfrac{\pi}{4}\right)=-\cos\dfrac{\pi}{4}=-\dfrac{1}{\sqrt2}, and sin⁡3π4=sin⁡(π−π4)=sin⁡π4=12\sin\dfrac{3\pi}{4}=\sin\left(\pi-\dfrac{\pi}{4}\right)=\sin\dfrac{\pi}{4}=\dfrac{1}{\sqrt2}. So z=2(−12+12i)=−1+iz=\sqrt2\left(-\dfrac{1}{\sqrt2}+\dfrac{1}{\sqrt2}i\right)=-1+i. Worked Example: express (i) 3e5πi/12×4eπi/123e^{5\pi i/12}\times4e^{\pi i/12} and (ii) 2(cos⁡5π6+isin⁡5π6)2(cos⁡π12+isin⁡π12)\dfrac{\sqrt2\left(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right)}{2\left(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\right)} in a+iba+ib form.

(i) 3e5πi/12×4eπi/12=(3×4)ei(5π12+π12)=12e6πi/12=12eiπ/2=12(cos⁡π2+isin⁡π2)=12(0+i)=12i3e^{5\pi i/12}\times4e^{\pi i/12}=(3\times4)e^{i\left(\frac{5\pi}{12}+\frac{\pi}{12}\right)}=12e^{6\pi i/12}=12e^{i\pi/2}=12\left(\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}\right)=12(0+i)=12i. …