CIE A Level Maths: Pure 3

Revision Notes

8.3.1 Exponential Form of Complex Numbers

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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.

How do we calculate e to the power of an imaginary number?

  • Given an imaginary number (iθ) we can define exponentiation as
    • straight e to the power of straight i theta end exponent equals cos space theta plus isin space theta
    • straight e to the power of straight i theta end exponent is the complex number with modulus 1 and argument θ
  • This works with our current rules of exponents
    • straight e to the power of 0 equals straight e to the power of 0 straight i end exponent equals cos invisible function application 0 plus isin invisible function application 0 equals 1
      • This shows e to the power 0 would still give the answer of 1
    • straight e to the power of straight i theta subscript 1 end exponent cross times straight e to the power of straight i theta subscript 2 end exponent equals straight e to the power of straight i left parenthesis theta subscript 1 plus theta subscript 2 right parenthesis end exponent 
      • 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
    • straight e to the power of iθ subscript 1 end exponent over straight e to the power of iθ subscript 2 end exponent equals straight e to the power of straight i left parenthesis straight theta subscript 1 minus straight theta subscript 2 right parenthesis end exponent 
      • 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 equals a plus b straight i can be written in polar form z equals r open parentheses cos invisible function application theta plus isin invisible function application theta close parentheses
    • r is the modulus
    • θ is the argument
  • Using the definition of straight e to the power of straight i theta end exponent we can now also write z in exponential form
    • z equals r straight e to the power of straight i theta end exponent

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 subscript 1 equals r subscript 1 straight e to the power of straight i theta subscript 1 end exponent and z subscript 2 equals r subscript 2 straight e to the power of straight i theta subscript 2 end exponent then
    • z subscript 1 cross times z subscript 2 equals r subscript 1 r subscript 2 straight e to the power of straight i open parentheses theta subscript 1 plus theta subscript 2 close parentheses end exponent
      • You can clearly see that the moduli have been multiplied and the arguments have been added
    • z subscript 1 over z subscript 2 equals r subscript 1 over r subscript 2 straight e to the power of straight i open parentheses theta subscript 1 minus theta subscript 2 close parentheses end exponent 
      • 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 space left parenthesis 2 pi right parenthesis equals 1 and sin space left parenthesis 2 pi right parenthesis equals 0 you can write:
    • 1 equals straight e to the power of 2 pi straight i end exponent
  • Using the same idea you can write:
    • 1 equals straight e to the power of 0 equals straight e to the power of 2 pi straight i end exponent equals straight e to the power of 4 pi straight i end exponent equals straight e to the power of 6 pi straight i end exponent equals straight e to the power of 2 k pi straight i end exponent where k is any integer
  • As cos invisible function application open parentheses pi close parentheses equals negative 1 and sin invisible function application left parenthesis pi right parenthesis equals 0 you can write:
    • straight e to the power of pi straight i end exponent equals negative 1
    • Or more commonly written as straight e to the power of iπ plus 1 equals 0
  • As cos invisible function application open parentheses pi over 2 close parentheses equals 0 and sin invisible function application open parentheses pi over 2 close parentheses equals 1 you can write:
    • straight i equals straight e to the power of pi over 2 straight i end exponent

Worked example

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Exam Tip

  • The powers can be long and contain fractions so make sure you write the expression clearly.
  • You don’t want to lose marks because the examiner can’t read your answer

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Dan

Author: Dan

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.