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Math 389 Fall 2012, Test 1, Oct. 16

Solutions

(1) (a) The given number is

1 + 2i

−8 + 6i  =  1

100(1 + 2i)(−8 − 6i) =   1

100(4 − 22i) =   1

25 − 11

50i

(b) 1 +√ 

3i = 2eiπ/3 so the given number is

251ei51π/3 = 251ei17π = −251.

(2) (a) 1 + i =√ 

2eiπ/4 so one cube roots is  z 0  = 21/6eiπ/12. Then every cube root is of the

form z 0ei2πk/3, k = 0, 1, 2 so we obtain 21/6ei(π/12)+k2π/3. (Remark: the result of 4(a)

could be used to write each of these out in the form  a  + bi.)(b) Since 1 + i = exp(log 2 + i(π/4 + 2kπ)), k

 ∈Z we have

(1 + i)i = exp(i log2 − (π/4 + 2kπ)), k ∈ Z.

(3) Translating everything by −c  the properties of being equilateral and being a Gaussianinteger do not change. Thus, we may assume that   c   = 0. Then if the triangle isequilateral a, b have equal lengths and the angle between them is  π/3, so (interchanginga, b if needed) we have  b  =  eiπ/3a = (1/2 + i

√ 3/2)a. If  a  =  m + in and  b  =  m′ + in′ then

it follows that  m′ =  m/2 − n√ 3/2 and  n′ =  n/2 +  m√ 3/2. If  n = 0 solving the firstequality for

√ 3 implies

√ 3 is rational (since  m, n, m′, n′ are integers), a contradiction. If 

n = 0 then  m = 0, since  a = 0 and the second equality leads to the same contradiction.Thus the triangle cannot be equilateral.

(4) (a)

eiπ/12 = eiπ/3−iπ/4 = eiπ/3e−iπ/4 =  1

2√ 

2(1 + i

√ 3)(1 + i) =

 1 − √ 32√ 

2+ i

1 +√ 

3

2√ 

2

(b) We can take  f (reit) = log r +  it   with −π/4   < t <   7π/4. Calling the domain   D,multiplying by   e−i3π/4 rotates  D  onto the domain of Log.   f (z ) and Log(e−i3π/4z )have the same real part (log |z |) and the imaginary parts differ by 3π/4 so we seethat f (z ) = Log(e−i3π/4z ) + i3π/4.

(5) (a) The interior of   E   consists of all those points   z  ∈   C   such that   D(z, r) ⊂   E   forsome   r >   0. The closure of   E   consists of all   z  ∈   C   such that for every   r >   0,D(z, r) ∩ E  = ∅. The boundary of  E  is the set of all  z  in the closure of  E  which arenot in the interior of  E .

(b) The interior of  E   is  E \ {−(1 + i)/2, (1 + i)/2}. The closure of  E   is {z   : |z | ≤  1}.The boundary of  E   is {z  : |z | = 1}∪ [−(1 + i)/2, (1 + i)/2], where the notation [a, b]means the closed line segment from  a  to  b  for any  a, b ∈ C.There was a typo in the definition of  F : [−3π/4, π/4] should have been [−3π/4, −π/4].Of course if you answered the question correctly as stated you will get full marks.

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The solution here is for  F   = {z  = 0 : Arg z  ∈ [ −3π4   ,   −π4   ] ∪ [π4 ,  3π4  ]}. The interior of  F is

{z  = 0 : Arg z  ∈ ( −3π4

  ,−π

4  ) ∪ ( π

4, 3π

4 )}.

The closure of  F   is just F  ∪{0}. The boundary is the union of the two lines passingthrough 0,   1 + i  and through 0, 1

−i.

(6)   f (z ) = z (1 − z 2)3 has the primitive  F (z ) = −(1−z2)48

  in  C so the integral over any closedpath is 0. For the same reason, the integral can be evaluated as  F (z 1) − F (z 0) wherez 0 =  γ (0) = 0 and  z 1  =  γ (1) = i. Thus 

γ 

f (z )dz  =  F (i) − F (0) = −2 + 18

 = −158

 .

(7) We have  a + b = −(c + d) so taking moduli and squaring we get |a + b|2 = |c + d|2. Nowthe left hand side of this last equation is

(a + b)(a + b) = |a|2

+ ab + ab + |b|2

= 1 + 2Re ab + 1.Expanding the right hand side in the same way and equating we get Re ab   = Re cd.Similarly

(0.1)   |a − b|2 = 2 − 2Re(ab) = 2 − 2Re(cd) = |c − d|2.If we assume that  a,b,c,d  occur in that order around the circle equation 0.1 says thattwo opposite sides of the quadrilateral have equal length.

Now the hypotheses of the problem remain the same if  a, b, c, d  are permuted amongeach other so equation 0.1 also implies that |a − c|2 = |b − d|2 and |a − d|2 = |b − c|2.Together the three equations say that both pairs of opposite sides have the same lengthand that the diagonals have the same length. It is clear that these three facts imply that

we have a rectangle. To actually prove this is a little exercise in Euclidean geometry(which you were not required to do): in a nutshell, use several pairs of congruent trianglesin the picture to show that the angles at the four vertices are equal, and hence each isπ/2.