Math 412 Group Work Solutions - Complex Numbers and Powers - Prof. Scott Annin, Assignments of Mathematics

The solutions to group work #10 in math 412 for spring 2009. The problems involve computing sets of complex numbers and finding principal values, proving identities, and solving for complex numbers. The document sheds light on the behavior of complex numbers when raised to complex powers and the relationship between exponentials and logarithms.

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Pre 2010

Uploaded on 08/16/2009

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Math 412 Group Work #10 Spring 2009
SOLUTIONS
Problem 1. Compute the sets below, and then find the principal values:
(a): (i2)i
SOLUTION: We have
(i2)i= (1)i=eilog(1) =ei(i(π+2πn)) =e(π+2πn).
The principal value is eπ.
(b): i2i
SOLUTION: We have
i2i=e2ilog i=e2i(i(π
2+2πn)) =e(π+4πn).
The principal value is eπ.
What do your calculations above show about what can happen when you
raise a complex number to a complex power?
SOLUTION: When you raise a complex number to a complex power, you can get
a real value!
Problem 2. Prove that ez=1
ezfor all zC.
SOLUTION: Since
1 = e0=ez+(z)=ezez,
we conclude that
ez=1
ez.
Problem 3. Use Problem 2 to show that if cCis fixed and zC, then
zc=1
zc.
SOLUTION: We have
zc= exp[c·log(z)] = 1
exp[c·log(z)] =1
zc,
where we used Problem 2 in the middle step.
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Math 412 Group Work #10 Spring 2009 SOLUTIONS

Problem 1. Compute the sets below, and then find the principal values:

(a): (i^2 )i

SOLUTION: We have

(i^2 )i^ = (−1)i^ = ei^ log(−1)^ = ei(i(π+2πn))^ = e−(π+2πn).

The principal value is e−π. 

(b): i^2 i

SOLUTION: We have

i^2 i^ = e^2 i^ log^ i^ = e^2 i(i(^

π 2 +2πn)) = e−(π+4πn).

The principal value is e−π. 

What do your calculations above show about what can happen when you raise a complex number to a complex power?

SOLUTION: When you raise a complex number to a complex power, you can get a real value! 

Problem 2. Prove that e−z^ =

ez^

for all z ∈ C.

SOLUTION: Since 1 = e^0 = ez+(−z)^ = ez^ e−z^ ,

we conclude that

e−z^ =

ez^

Problem 3. Use Problem 2 to show that if c ∈ C is fixed and z ∈ C∗, then

z−c^ =

zc^

SOLUTION: We have

z−c^ = exp[−c · log(z)] =

exp[c · log(z)]

zc^

where we used Problem 2 in the middle step. 

Problem 4. Solve for z: exp(z + 1) = i.

SOLUTION: Taking the logarithm of each side, we have

z + 1 = log(i) = {ln |i| + i(

π 2

  • 2πn) : n ∈ Z} = {i(

π 2

  • 2πn) : n ∈ Z}.

Thus, z can be any element of the set

{−1 + i(

π 2

  • 2πn) : n ∈ Z}.