1. Complex Numbers – Part 2 By Dr. Samer Awad
Assistant professor of biomedical engineering
The Hashemite University, Zarqa, Jordan
Last update: 27 December 2018

2. 13.5 Exponential Function 𝑒 𝑧2
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13.5 Exponential Function 𝑒 𝑧
• 𝑧=𝑥+𝑖𝑦
• 𝑒 𝑧 = 𝑒 𝑥+𝑖𝑦 = 𝑒 𝑥 𝑒 𝑖𝑦
• The complex exponential function ez can be expressed as:
𝑒 𝑧 =𝑒𝑥 (cos𝑦+𝑖 𝑠𝑖𝑛𝑦)
 𝑒 𝑖𝑦 = cos𝑦+𝑖 𝑠𝑖𝑛𝑦 , but how is that?

3. Exponential Function 𝑒 𝑧3
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Exponential Function 𝑒 𝑧
Using Taylor series expansions we need to prove that :
Euler’s formula.

4. Exponential Function 𝑒 𝑧4
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Exponential Function 𝑒 𝑧
• The complex exponential function ez can be expressed as:
𝑒 𝑧 =𝑒𝑥(cos𝑦+𝑖 𝑠𝑖𝑛𝑦)
• Also, since 𝑧 = 𝑟(cos𝜃+𝑖 𝑠𝑖𝑛𝜃), the polar form of a complex number z can be expressed as:
𝑧 = 𝑟(cos𝜃+𝑖 𝑠𝑖𝑛𝜃)
𝑧 = 𝑟 𝑒 𝑖𝜃

5. Properties of 𝑒 𝑧 • ( 𝑒 𝑧 )′= 𝑒 𝑧 ( 𝑒 𝑧 )′= 𝑑 𝑑𝑥 ( 𝑒 𝑧 )5
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Properties of 𝑒 𝑧
• ( 𝑒 𝑧 )′= 𝑒 𝑧 ( 𝑒 𝑧 )′= 𝑑 𝑑𝑥 ( 𝑒 𝑧 )
• 𝑒 𝑧1 𝑒 𝑧2 =𝑒 𝑧1+𝑧2
• 𝑒 𝑖2𝜋 =1, but why?
• Also, 𝑒 𝑖𝜋/2 =𝑖, 𝑒 𝑖𝜋 =−1, 𝑒 −𝑖𝜋/2 =−𝑖
• How can you plot these on the complex plane?

6. 6
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Properties of 𝑒 𝑧
• 𝑒𝑖𝑦 = cos𝑦+𝑖 𝑠𝑖𝑛𝑦 = cos2𝑦+ 𝑠𝑖𝑛2𝑦 =1 pure imaginary part amplitude=1.
• Hence, 𝑒𝑧 = 𝑒 𝑥 and arg 𝑒𝑧=𝑦±𝑛2𝜋
• 𝑒 𝑧 is periodic with period = 𝑖2𝜋
– 𝑒 𝑧+𝑖𝑛2𝜋 = 𝑒 𝑧 .
– All values for w= 𝑒 𝑧 are
within a region = 2𝜋.
– The fundamental region
of 𝑒 𝑧 is: −𝜋<𝑦≤𝜋.

7. Examples: Exponential Function7
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Examples: Exponential Function
• Find the Cartesian & polar form of 𝑒 1.4−0.6𝑖
• To prove that 𝑒 𝑧1 𝑒 𝑧2 = 𝑒 𝑧1+𝑧2 for:
• Solve 𝑒 𝑧 =3+4𝑖:

8. Trigonometric Functions8
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Trigonometric Functions
• Euler’s formula:
𝑒 𝑖𝑥 =cos𝑥+𝑖 𝑠𝑖𝑛𝑥… 𝑒𝑞 𝑒 −𝑖𝑥 =cos𝑥−𝑖 𝑠𝑖𝑛𝑥…(𝑒𝑞2)
• Add the previous two eq’s (eq1+eq2) to get:
𝑒 𝑖𝑥 + 𝑒 −𝑖𝑥 =2 cos 𝑥  cos 𝑥 = 𝑒 𝑖𝑥 + 𝑒 −𝑖𝑥
• Subtract those eq’s (eq1 – eq2) to get:
𝑒 𝑖𝑥 − 𝑒 −𝑖𝑥 =2𝑖 sin 𝑥  sin 𝑥 = 1 2𝑖 𝑒 𝑖𝑥 − 𝑒 −𝑖𝑥
• Similarly, for a complex value 𝑧=𝑥+𝑖𝑦:
cos 𝑧 = 𝑒 𝑖𝑧 + 𝑒 −𝑖𝑧 sin 𝑧 = 1 2𝑖 𝑒 𝑖𝑧 − 𝑒 −𝑖𝑧

9. Trigonometric Functions9
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Trigonometric Functions
• The other trigonometric functions are defined as:
tan 𝑧= sin 𝑧 cos 𝑧 cot 𝑧= cos 𝑧 sin 𝑧
sec 𝑧= 1 cos 𝑧 csc 𝑧= 1 sin 𝑧

10. sinh 𝑧 = 1 2 𝑒 𝑧 − 𝑒 −𝑧 cosh 𝑧 = 1 2 𝑒 𝑧 + 𝑒 −𝑧10
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Hyperbolic Functions
sinh 𝑧 = 𝑒 𝑧 − 𝑒 −𝑧 cosh 𝑧 = 𝑒 𝑧 + 𝑒 −𝑧

11. 11
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Hyperbolic Functions
• The complex hyperbolic cosine and sine are defined by the formulas:
sinh 𝑧 = 𝑒 𝑧 − 𝑒 −𝑧 cosh 𝑧 = 𝑒 𝑧 + 𝑒 −𝑧
• Complex Trigonometric and Hyperbolic Functions Are Related:
cosh 𝑖𝑧 = 𝑒 𝑖𝑧 + 𝑒 −𝑖𝑧 = cos 𝑧
sinh 𝑖𝑧= 𝑒 𝑖𝑧 − 𝑒 −𝑖𝑧 =𝑖 sin 𝑧
• Conversely:
cos 𝑖𝑧 = cosh 𝑧 sin 𝑖𝑧 =𝑖 sinh 𝑧

12. Hyperbolic Functions tanh 𝑧= sinh 𝑧 cosh 𝑧 cot 𝑧= cosh 𝑧 sinh 𝑧12
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Hyperbolic Functions
• The other trigonometric functions are defined as:
tanh 𝑧= sinh 𝑧 cosh 𝑧 cot 𝑧= cosh 𝑧 sinh 𝑧
sech 𝑧= 1 cosh 𝑧 csc 𝑧= 1 sinh 𝑧

13. Examples: Trigonometric & Hyperbolic Functions13
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Examples: Trigonometric & Hyperbolic Functions
Prove the following equation 6a

14. Examples: Trigonometric & Hyperbolic Functions14
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Examples: Trigonometric & Hyperbolic Functions
Prove the following equation 7a

15. Example: Trigonometric & Hyperbolic Functions15
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Example: Trigonometric & Hyperbolic Functions
• cos𝑧= 𝑒 𝑖𝑧 + 𝑒 −𝑖𝑧 𝑠𝑖𝑛𝑧= 1 2𝑖 𝑒 𝑖𝑧 − 𝑒 −𝑖𝑧 …(1)

16. Natural Logarithm Function16
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Natural Logarithm Function
• The natural logarithm of z is denoted by:
𝒘= 𝐥𝐧 𝒛 → 𝑒 𝑤 =𝑧 𝑓𝑜𝑟 𝑧≠0
• Let 𝑧=𝑟 𝑒 𝑖𝜃 :
ln 𝑧 = ln 𝑟 +𝑖 𝜃 𝑂𝑅 ln 𝑧 = ln 𝑟 +𝑖 arg 𝑧
• Since arg z corresponds to the same value every 2𝜋, ln 𝑧 has infinite values (multivalued).
• The value of ln z corresponding to the principal value Arg z is denoted by Ln 𝑧 :
Ln 𝑧 = ln 𝑟 +𝑖 Arg z

17. Natural Logarithm Function17
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Natural Logarithm Function
• Ln 𝑧 is called the principal value of ln 𝑧 :
Ln 𝑧 = ln |𝑧| +𝑖 Arg z
ln 𝑧 = Ln 𝑧 ±𝑖 𝑛2𝜋
• If z is positive real, then Arg z = 0 and Ln z becomes regular ln(x) function from calculus.
• If z is negative real, (remember ln(-x) is not defined in calculus):
Ln 𝑧 = ln |𝑧| +𝑖 Arg z = ln |𝑧| +𝑖 𝜋

18. Natural Logarithm Function18
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Natural Logarithm Function
• From: ln 𝑧 = ln 𝑟 +𝑖𝜃, it follows that:
𝑒 ln 𝑧 = 𝑒 ln 𝑟 𝑒 𝑖𝜃 =𝑧 (single-valued)
• Since arg 𝑒 𝑧 =𝑦±𝑛2𝜋 , it follows that:
ln 𝑒 𝑧 =𝑧±𝑖 𝑛2𝜋 (multi-valued)
Remember:
𝑒 𝑧 is periodic  𝑒 𝑧 = 𝑒 𝑧±𝒊 𝒏2𝜋
ln 𝑒 𝑧 = ln 𝑒 𝑧±𝑖 𝑛2𝜋 = ln 𝑒 𝑥+𝑖𝑦±𝑖 𝑛2𝜋 =𝑥+𝑖𝑦±𝑖 𝑛2𝜋
=𝑧±𝑖 𝑛2𝜋

19. Examples: Natural Logarithm19
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Examples: Natural Logarithm
• Solve in class

20. Examples: Natural Logarithm20
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Examples: Natural Logarithm

21. Natural Logarithm Function21
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Natural Logarithm Function
• The familiar relations for the natural logarithm continue to hold for complex values:
ln (𝑧1𝑧2) = ln (𝑧1) + ln (𝑧2)
ln (𝑧1/𝑧2) = ln (𝑧1) − ln (𝑧2)

22. Example: Natural Logarithm22
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Example: Natural Logarithm
• Prove that ln (𝑧1𝑧2) = ln (𝑧1) + ln (𝑧2) and that
Ln (𝑧1𝑧2) ≠ Ln (𝑧1) + Ln (𝑧2) for z1 = z2 = -1

23. General Powers • 𝑧=𝑥+𝑖𝑦 • 𝑧 𝑐 = 𝑒 ln 𝑧 𝑐 = 𝑒 𝑐 ln 𝑧 𝑐 𝑐𝑜𝑚𝑝𝑙𝑒𝑥, 𝑧≠0 .23
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General Powers
• 𝑧=𝑥+𝑖𝑦
• 𝑧 𝑐 = 𝑒 ln 𝑧 𝑐 = 𝑒 𝑐 ln 𝑧 𝑐 𝑐𝑜𝑚𝑝𝑙𝑒𝑥, 𝑧≠0 .
• ln z is multivalued  𝑧 𝑐 is multivalued.
• Hence, there is a principal value of zc which is:
𝑧 𝑐 =𝑒 c Ln 𝑧

24. General Powers • 𝑧 𝑐 = 𝑒 𝑐 ln 𝑧 𝑐 𝑐𝑜𝑚𝑝𝑙𝑒𝑥, 𝑧≠0 .24
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General Powers
• 𝑧 𝑐 = 𝑒 𝑐 ln 𝑧 𝑐 𝑐𝑜𝑚𝑝𝑙𝑒𝑥, 𝑧≠0 .
• If c = n = 1, 2, … then 𝑧 𝑛 is single-valued and identical with the usual nth power of z.
• If c = n = -1, -2, … then 𝑧 𝑛 is also single-valued .
• If c = 1/n, where n = 2, 3, … , then:
𝑧 𝑐 = 𝑧 1/𝑛 = 𝑛 𝑧
 Same finite n roots explained previously.

25. General Powers • 𝑧 𝑐 = 𝑒 𝑐 ln 𝑧 𝑐 𝑐𝑜𝑚𝑝𝑙𝑒𝑥, 𝑧≠0 .25
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General Powers
• 𝑧 𝑐 = 𝑒 𝑐 ln 𝑧 𝑐 𝑐𝑜𝑚𝑝𝑙𝑒𝑥, 𝑧≠0 .
• If c=m/n, the quotient of two positive integers, the situation is similar, and 𝑧 𝑐 has only finite n distinct values (n roots).
• However, if c is real irrational or complex, then 𝑧 𝑐 is infinitely many-valued.
• Remember: an irrational number is any real number that cannot be expressed as a ratio of integers. Example: 𝜋, 2 .

26. Example: General Powers26
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Example: General Powers
• Find the values of: 𝑖 𝑖 , (1+𝑖) 2−𝑖 :