Exponential Form of Complex Numbers (Cambridge (CIE) A Level Maths: Pure 3): Revision Note

Exponential form of complex numbers

You now know how to do lots of operations with complex numbers: add, subtract, multiply, divide, raise to a power and even square root. The last operation to learn is raising the number e to the power of an imaginary number.

What is e to the power of an imaginary number?

• Given an imaginary number (iθ) we can define exponentiation as

  • ${\mathrm{e}}^{\mathrm{i}\theta }=\cos \theta +\mathrm{isin}\theta$

  • ${\mathrm{e}}^{\mathrm{i}\theta }$ is the complex number with modulus 1 and argument θ

• This works with our current rules of exponents

  • ${\mathrm{e}}^0={\mathrm{e}}^{0\mathrm{i}}=\cos 0+\mathrm{isin}0=1$

    • This shows e to the power 0 would still give the answer of 1

  • ${\mathrm{e}}^{\mathrm{i}{\theta }_1}\times {\mathrm{e}}^{\mathrm{i}{\theta }_2}={\mathrm{e}}^{\mathrm{i}({\theta }_1+{\theta }_2)}$ 

    • This is because when you multiply complex numbers you can add the arguments

    • This shows that when you multiply two powers you can still add the indices

  • $\frac{{\mathrm{e}}^{{\mathrm{i\theta }}_1}}{{\mathrm{e}}^{{\mathrm{i\theta }}_2}}={\mathrm{e}}^{\mathrm{i}({\mathrm{\theta }}_1-{\mathrm{\theta }}_2)}$ 

    • This is because when you divide complex numbers you can subtract the arguments

    • This shows that when you divide two powers you can still subtract the indices

What is the exponential form of a complex number?

• Any complex number $z=a+b\mathrm{i}$ can be written in polar form $z=r\left(\cos \theta +\mathrm{isin}\theta \right)$

  • r is the modulus

  • θ is the argument

• Using the definition of ${\mathrm{e}}^{\mathrm{i}\theta }$ we can now also write $z$ in exponential form

  • $z=r{\mathrm{e}}^{\mathrm{i}\theta }$

Why do I need to use the exponential form of a complex number?

• It's just a shorter and quicker way of expressing complex numbers

• It makes a link between the exponential function and trigonometric functions

• It makes it easier to remember what happens with the moduli and arguments when multiplying and dividing

• If $z_1=r_1{\mathrm{e}}^{\mathrm{i}{\theta }_1}$ and $z_2=r_2{\mathrm{e}}^{\mathrm{i}{\theta }_2}$ then

  • $z_1\times z_2=r_1{r}_2{\mathrm{e}}^{\mathrm{i}\left({\theta }_1+{\theta }_2\right)}$

    • You can clearly see that the moduli have been multiplied and the arguments have been added

  • $\frac{z_1}{z_2}=\frac{r_1}{r_2}{\mathrm{e}}^{\mathrm{i}\left({\theta }_1-{\theta }_2\right)}$ 

    • You can clearly see that the moduli have been divided and the arguments have been subtracted

What are some common numbers in exponential form?

• As $\cos \left(2\pi \right)=1$ and $\sin \left(2\pi \right)=0$ you can write:

  • $1={\mathrm{e}}^{2\pi \mathrm{i}}$

• Using the same idea you can write:

  • $1={\mathrm{e}}^0={\mathrm{e}}^{2\pi \mathrm{i}}={\mathrm{e}}^{4\pi \mathrm{i}}={\mathrm{e}}^{6\pi \mathrm{i}}={\mathrm{e}}^{2k\pi \mathrm{i}}$ where k is any integer

• As $\cos \left(\pi \right)=-1$ and $\sin \left(\pi \right)=0$ you can write:

  • ${\mathrm{e}}^{\pi \mathrm{i}}=-1$

  • Or more commonly written as ${\mathrm{e}}^{\mathrm{i\pi }}+1=0$

• As $\cos \left(\frac{\pi }{2}\right)=0$ and $\sin \left(\frac{\pi }{2}\right)=1$ you can write:

  • $\mathrm{i}={\mathrm{e}}^{\frac{\pi }{2}\mathrm{i}}$