Problems about Complex numbers

• READ THIS FIRST
  
• Problems about Complex numbers
  
  • Level 1 problems
    
  • Level 2 problems
    
  • Level 3 problems
    

READ THIS FIRST

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No attempt is made to search for the most elegant answer.
I highly recommend that you at least try to solve the problem before you read the solution.

Problems about Complex numbers

Level 1 problems

• 
  

   
  
  (4i-7)+(1-i) = 3i-6
  
  -(5-i) = -5+i
  
  i.(i-1) = -1-i
  
   1
  --- = -i
   i
  
  |-1| = 1
  
  |-i| = 1
  
  |-4| = 4
  

• 
  

  1+i --- = ? 1-i

  
  

   
  (1+i)(1+i)
  ----------- =
  (1-i)(1+i)
  
  1+2i-1
  ------- = i
    1+1
  
  

• 
  

  i2012 = ?

  
  

   
  i4 = 1
  
  So
   4n
  i   = 1  (n is a positive integer)
  
  i2012 = 1
  

• 
  

  solve z2 = -4i

  
  

   
  Let x+iy  a square root of -4i.
   
  The modulus of -i is  r = 4
   
  y = sqrt((a + r)/2) = sqrt(2)
   
  x =  b/(2.y) = (-4)/(2.sqrt(2)) = -sqrt(2)
   
  The two solutions are 
   
  -sqrt(2)+i sqrt (2)
   
  +sqrt(2)-i sqrt(2)
  
   
  

• 
  

  Show that for each complex number z _ z . z = a real number

  
  

   
  
  Say z = a+ib, then
  _
  z . z = (a+bi)(a-bi) = a2 + b2 = real number
  

• 
  

  Calculate the conjugate complex number of z = a + bi 2 a - bi 2 (--------) + (--------) a - bi a + bi

  
  

   
  
  Since
   conj(c.c') = conj(c).conj(c')
   conj(c/c') = conj(c)/conj(c')
   conj(c + c') = conj(c) + conj(c')
  
  the conjugate is
    a - bi  2     a + bi  2
  (--------)  + (--------) = z
    a + bi        a - bi
  

  Since the given complex number is equal to its conjugate, it is a real number.
  
• 
  

  Solve : ix2 +(1-5i)x -1+8i=0

  
  

   
  Discriminant = ... = -6i+8
  The two square roots out of -6i+8 are 3-i and -3+i
  The roots of the given equation are then
  2-i and 3+2i
  

• 
  

  Find the polar representation of (i-sqrt(3))

  
  

   
  The modulus of (i-sqrt(3)) is 2.
   (i-sqrt(3)) = 2.( -sqrt(3)/2 + (1/2)i )
  Say, the argument is alpha.
  cos(alpha) = -sqrt(3)/2
  sin(alpha) = (1/2)
  Choose alpha = 5 pi/6
  (i-sqrt(3)) = 2.(cos(5.pi/6) + i sin(5.pi/6))
  
  

• 
  

  Simple calculations

  
  

   
  2.(cos(1) +i sin(1)).5.(cos(2) +i sin(2))= 10.(cos(3) +i sin(3))
  
  6.(cos(5) +i sin(5))
  --------------------= 2.(cos(3) +i sin(3))
  3.(cos(2) +i sin(2))
  
  
  (2.(cos(3) +i sin(3)))5  =  32.(cos(15) +i sin(15))
  
  

• 
  

  Find all z so that z4 = -8(i-sqrt(3))

  
  

   
  
  -8(i-sqrt(3)) = 16.(cos(-pi/6) + i sin(-pi/6))
  The 4th roots are
  z = 2.(cos(-pi/24 + k.pi/2) + i sin(-pi/24 + k.pi/2))  with k in Z
  
  

• 
  

  Given : z=cos(3)+ i sin(3) _ Prove that 1 + z = (1 + z )z

  
  

   
  
       _           _
  (1 + z )z = (z + z.z) = cos(3)+ i sin(3) + (cos(3)+ i sin(3))(cos(3) - i sin(3))
  
  = cos(3)+ i sin(3) + 1 = 1 + z
  
  

• 
  

  Given : u = 1+i.sqrt(3) and v = sqrt(3) + i Calculate u3 / v4

  
  
  With the polar form
  u = 2 ( cos(pi/3) + i sin(pi/3) ) en v = 2 ( cos(pi/6) + i sin(pi/6) )

  Then u³ = 8(cos(pi) + i sin(pi)) = -8 and v⁴ = 16( cos(2 pi/3) + i sin(2 pi/3) )

  u³ / v⁴ = -(1/2)( cos(2 pi/3) - i sin(2 pi/3) ) = (1/4) - i sqrt(3)/4

• 
  

  Show that the equation has a real root. 4z3 - 6i sqrt(3) z2 - 3(3 + i sqrt(3)) z - 4 = 0

  
  
  If there exists a real root w then the imaginary part of the left hand side must be 0 for z = w.
  - 6i sqrt(3) w² - 3.i sqrt(3) w = 0.
  This is equivalent to 2 w² + w = 0. So, w = 0 or w = -1/2
  But w = 0 can not be a solution of the given equation.
  If there exists a real root w then it is -1/2.
  Now we check if -1/2 is a root of the given equation and we see that it is so.
• 
  

  Find (1+i)17 --------- (1-i)16

  
  

   
      (1+i)
     -------- = i
      (1-i)
  So
      (1+i)16
      --------- = 1
      (1-i)16
  and
      (1+i)17
      --------- = 1+i
      (1-i)16
  

Level 2 problems

• 
  

  Given: z not real and |z|= 1 z-1 Show that w = --- is a pure imaginary number. z+1

  
  

   
  
  Let z = a+bi.
  
        a+bi-1    (a+bi-1)(a-bi+1)    (a-1+bi)(a+1-bi)    a.a+b.b-1 + 2bi
   w =  ------  =  ---------------- = -----------------= -----------------
        a+bi+1    (a+bi+1)(a-bi+1)    (a+bi+1)(a-bi+1)    (a+1)(a+1)+b.b
  
  but a2 + b2 = 1, so
  
             2bi
   w = ---------------
       (a+1)(a+1)+b2
  

  From this expression it is obvious that w is a pure imaginary number.
• 
  

  Prove that in C, there are no divisors of zero. That is, z.z'=0 => (z=0 or z'=0)

  
  

   
  If z=0, the statement is proved.
  
   
  
  If z not 0, then there is a complex number z" (not zero) so that z".z=1.
   
  Then, z.z'=0 => z".z.z'=z".0 => 1.z'=0 => z'=0
   
  

• 
  

  Calculate ( cos(2)+ i sin(2) + 1)n

  
  

   
  Appealing on trigonometric formulas we have
  
  (1 + cos(2)) = 2 cos2 (1)  and sin(2)=2.sin(1).cos(1)
  
  (1 + cos(2)+ i sin(2)) =  2 cos2 (1) + i.2.sin(1).cos(1)
  
                        = 2.cos(1) (cos(1) + i sin(1))
  
  (1 + cos(2)+ i sin(2))n  = 2n .cosn (1) .(cos(n) + i sin(n))
  

• 
  

  The image point of z = a + bi in the Gauss-plane is p. We rotate p about o and the angle of the rotation is pi/3. The new position of p is p'. Calculate the coordinates of p'.

  
  

   
  Say z has polar notation r(cos(t)+i sin(t))
  
  p' is the image point of z.(cos(pi/3) + i sin(pi/3)) in the Gauss-plane
  
  Hence,  p' is the image point of r(cos(t+pi/3) + i sin(t+pi/3))
  
  The coordinates of p' are (r.cos(t+pi/3) ; r.sin(t+pi/3))
  

• 
  

  a, b, c are real numbers in the polynomial p(z) = 2 z4 + a z3 + b z2 + c z + 3 . Find a such that the numbers 2 and i are roots of p(z) = 0.

  
  

   
  Since all the coefficients of p(z) are real, -i is a root of p(z) = 0.
  
  Let 2, i, -i, w be all the roots.
  
  The sum of the roots  = 2 + w = -a/2.
  
  The product of the roots = 2w = 3/2.
  
  From this we find w = 3/4 and a = -11/2.
  

• 
  

  Given: n is a positive integer. z is a complex number with modulus 1, such that z2n is not -1. zn Show that -------- is a real number 1 + z2n

  
  

   
  
  
  Since z is a complex number with modulus 1, we can write
  
          z = (cos(t) + i sin(t))
  
          zn = (cos(n t) + i sin(n t))
  
          1 + z2n   = 1 + cos(2n t) + i sin(2n t)
  
                  = 2 cos2 (n t) + 2 i sin(n t) cos(n t)
  
                  = 2 cos(n t). (cos(n t) + i sin(n t))
  
  
              zn            1
           --------  = ------------   and this is real
           1 + z2n     2 cos(n t)
  
  

• 
  

  Calculate all integers n such that zn = (1 + i sqrt(3))n is a real number.

  
  
  z₁ = (1 + i sqrt(3)) has modulus 2 and argument = pi/3.
  Thus, z_n has modulus 2ⁿ and argument n.pi/3.
  z_n is real if and only if the argument is k.pi (with k = integer).
  So, z_n is real if and only if n is a multiple of 3.
• 
  

  Calculate the real values of x and y such that (x + iy)3 is real and |x + i y| is higher than 8.

  
  

   
     (x + iy)3 = x3 + 3 i x2 y - 3 x y2 - i y3
  
     (x + iy)3 is real
  <=>
     3 x2 y - y3 = 0
  <=>
     y (3 x2 - y2 ) = 0
  <=>
     y = 0 or 3 x2 - y2 = 0
  <=>
     y = 0 or 3 x2 = y2
  <=>
     y = 0 or y = sqrt(3) x or y = - sqrt(3) x
  

  Conclusion: |x + iy| > 8 if and only if

  1. 
     

      
             ( x > 2  and y = 0 )
     

  2. 
     

      
             y = sqrt(3) x  and  x2 + y2 > 64
     <=>
             y = sqrt(3) x  and   4x2 > 64
     <=>
             y = sqrt(3) x  and   x2 > 16
     <=>
             y = sqrt(3) x  and   |x| > 4
     

  3. 
     

      
             y = - sqrt(3) x  and  x2 + y2 > 64
     <=>
             y = - sqrt(3) x  and   4x2 > 64
     <=>
             y = - sqrt(3) x  and   x2 > 16
     <=>
             y = - sqrt(3) x  and   |x| > 4
     

• 
  

  Given: complex number z = cos(2t) + i sin(2t) Show that 2/(1+z) = 1 - i tan(t)

  
  

   
  1+ z = 1 + cos(2t) + i sin(2t)
  
  
       = 2 cos2(t) + i sin(2t)
  
  
              2            2 tan(t)
       = ---------- + i -------------
         1 + tan2(t)     1 + tan2(t)
  
  
             2 (1  + i tan(t))
       = ----------------------
             1 + tan2(t)
  
  
  
                2
       = ---------------
           1 - i tan(t)
  

• 
  

  Find the real value of m such that the equation 2 z2 - ( 3+ 8i )z - ( m + 4i) = 0 has a real root. Then find the roots.

  
  

  Say t is the real root. Then

   
     2 t2 - ( 3+ 8i )t - ( m + 4i) = 0
  <=>
     / 2t2 - 3t - m = 0
     \  -8t - 4 = 0
  <=>
     ...
  <=>
     t = -1/2  ; m = 2
  

  The real root is -1/2. The product of the roots is ( -1 - 2i). The second root is 2 + 4i.

Level 3 problems

• 
  

  Find real values of the number a for which a.i is a solution of the polynomial equation z4 - 2z3 + 7z2 - 4z + 10 = 0. Then find all roots of this equation.

  
  
  Since a.i is a solution of the equation, we have

   
     (a.i)4 - 2(a.i)3 + 7(a.i)2 - 4(a.i) + 10 = 0
  <=>
      a4 + 2.i.a3 - 7a2 - 4.i.a + 10 = 0
  <=>
      a4 - 7a2 + 10 = 0  and 2a3 - 4a = 0
  <=>
      a2 = 2
  <=>
      a = sqrt(2) or a = - sqrt(2)
  

  Now, we know that sqrt(2).i and - sqrt(2).i are roots of z⁴ - 2z³ + 7z² - 4z + 10 = 0 .
  This means that z⁴ - 2z³ + 7z² - 4z + 10 is divisible by (z - sqrt(2).i)(z + sqrt(2).i) = z² - 2.

  The quotient is (z² - 2 z + 5) . The roots of this polynomial are 1 + 2 i and 1 - 2 i.

   
  The four roots of the given equation are
  sqrt(2).i     -sqrt(2).i     1 + 2 i    1 - 2 i.
  

• 
  

  u,v and w are the three roots of the equation z3 - 1 = 0 . Calculate u.v + v.w + w.u without calculating the 3 roots.

  
  
  The roots are 1 and two conjugate complex numbers.
  Say u = 1. Then we have to calculate v + w + v.w .
  Since the sum of the roots is zero, we have 1 + v + w = 0 . Hence v + w = -1.
  We have to calculate -1 + v.w .
  Since the product of the roots is 1, we have 1.v.w = 1.
  So, u.v + v.w + w.u = -1 + v.w = -1 + 1 = 0
  

• 
  

  Calculate all solutions of |z-1|.|z-1|=1

  
  
  Let z = x + iy

   
          |z - 1|2  = 1
  <=>
          |x + iy - 1|2  = 1
  <=>
          (x - 1)2  + y2  = 1
  

  We see that the solutions are exactly the points (in the Gauss plane) of the circle with center (1,0) and radius 1.

• 
  

  The equation z3 - (n + i) z + m + 2 i = 0 has three roots. n and m are real constants. a) Calculate m such that the modulus of the product of the roots is 5. b) Calculate the modulus of the sum of the roots.

  
  

   
      |z1  z2  z3 | = 5 <=> |-(m+2i)|= 5 <=> ...<=> m = 1 or -1
  

  Since the sum of the roots is 0, the modulus of the sum is 0.

• 
  

  Let z' the conjugate complex number of z. Now find z such that z2 + z'2 = 0

  
  

   
  Let z = x + iy, then z' = x - iy
  
         z2  + z'2  = 0  <=> ... <=> 2 x2  - 2 y2  = 0
  
  <=> (x + y) (x - y) = 0 <=> y = x or y = -x
  

  So, z = x + ix or z = x - ix with arbitrary real x.
  In the Gauss plane, these solutions constitute the two bisector lines.

• 
  

  In the following equation, m is a real number. z2 - (3 + i) z + m + 2 i = 0 Calculate the values of m such that the equation has a real root. Calculate the second root.

  
  
  Say r is a real root. Then we have

   
        r2  - (3 + i) r + m + 2 i = 0
  
  <=>   r2  - 3 r + m + i(2 - r) = 0
  
  <=>   r2  - 3 r + m = 0 and 2 - r = 0
  
  <=>   r = 2  and m = 2
  

  For m = 2, there is a real root r = 2. But the sum of the roots is 3+i. From this, the second root is 1+i.

• 
  

  The number t is real and not an integer multiple of (pi/2). The complex numbers x1 and x2 are the roots of the equation tan2 (t).x2 + tan(t).x + 1 = 0 Show that (x1)n + (x2)n = 2 cos(2 n pi/3) cot(t)

  
  

   
  The discriminant of the quadratic equation is
  
          D = tan2 (t) - 4 tan2 (t) = -3 tan2 (t)
  
  The roots x1 and x2 are
  
                - tan(t) + i sqrt(3) tan (t)
          x1 = -------------------------------
                  2 tan (t) tan (t)
  
                -1 + sqrt(3)
             = ------------- cot(t)
                    2
  
  
             = (cos(2 pi/3) + i sin(2 pi/3)) cot(t)
  
  
                - tan(t) - i sqrt(3) tan (t)
          x2 = -------------------------------
                  2 tan (t) tan (t)
  
                -1 - sqrt(3)
             = ------------- cot(t)
                    2
  
  
             = (cos(2 pi/3) - i sin(2 pi/3)) cot(t)
  
  So,
  
          x1n  = (cos(2 pi/3) + i sin(2 pi/3))n  cotn(t)
  
             = (cos(2 n pi/3) + i sin(2 n pi/3))  cotn(t)
  
          x2n  = (cos(2 pi/3) - i sin(2 pi/3))n  cotn(t)
  
             = (cos(2 n pi/3) - i sin(2 n pi/3))  cotn(t)
  
  and
  
          (x1)n  + (x2)n  = 2 cos(2 n pi/3)  cotn(t)
  

• 
  

  Calculate the values of m such that the roots x1 and x2 of x2 - 2m x + m = 0 satisfy the condition x13 + x23 = x12 + x22. Calculate the roots for those m-values and check the condition.

  
  

   
  We see that s = x1 + x2 = 2m and p = x1 x2 = m.
  
  We write  x13 + x23 and x12 + x22 as a function of s and p.
  
  x13 + x23 = (x1 + x2)3 -3x1.x2(x1 + x2) = 8m3 -3.2m2
  
  x12 + x22 = (x1 + x2)2 - 2x1.x2 = 4m2 -2m
  
  Thus x13 + x23 = x12 + x22
  
  <=>  8m3 -3.2m2 =  4m2 -2m
  
  <=>  8m3 - 10m2 + 2m = 0
  
  <=>  m ( 4m2 - 5m + 1) = 0
  
  <=>  m = 0 or m = 1 or m = 1/4
  
  For m = 0 the roots are 0 and 0. The condition is satisfied.
  For m = 1 the roots are 1 and 1. The condition is satisfied.
  For m = 1/4 the equation is x2 - 1/2 x + 1/4 = 0
  
      The discriminant is -3/4
      The roots are x1 = (1 + sqrt(3)i)/4  and x2 = (1 - sqrt(3)i)/4
  To check the condition, we use the trigonometric form of the complex roots.
  
     x1 = 1/2 . (cos(pi/3) + i sin(pi/3))
  
     x12 = 1/4 .(cos(2 pi/3) + i sin(2 pi/3))
  
     x13 = - 1/8
  
     x2 = 1/2 . (cos(pi/3) - i sin(pi/3))
  
     x22 = 1/4 .(cos(2 pi/3) - i sin(2 pi/3))
  
     x23 = - 1/8
  
  x13 + x23 = -1/4 and x12 + x22 = 1/2 . cos(2 pi/3) = -1/4
  

• 
  

  Find all a-values such that the following statement is true. In C, the set of all roots of z8 - 1 = 0 is { ak | k in {1,2,3,4,5,6,7,8} }.

  
  

  We calculate all the roots of z⁸ = 1. These roots are the eighth roots of 1.

   
     1 = 1 ( cos(0 + 2kpi) + i sin(0 + 2kpi) )
  
       =  cos(2kpi) + i sin(2kpi)
  

  The eight roots are

   
        cos(2kpi/8) + i sin(2kpi/8) with k in  {0,1,2,3,4,5,6,7}
  

  It is clear that the number cos(2kpi/8) + i sin(2kpi/8) has the same value for k =0 as for k=8. The eight roots are also

   
        cos(2kpi/8) + i sin(2kpi/8) with  k in  {1,2,3,4,5,6,7,8}
  
    Let  b = cos(2 pi/8) + i sin(2 pi/8)
  
    The  eight roots are  { bk | k in {1,2,3,4,5,6,7,8} }.
  

  In the Gauss-plane the roots are the vertices of a regular octagon. The question to resolve here is: give the powers of b that generate all the vertices.
  They are the powers b^r such that r and 8 are coprime.
  The eight roots are generated by b, b³, b⁵, b⁷.
  So, we find 4 values of a namely a1=b ; a2=b³ a3=b⁵ and a4=b⁷.

  

• 
  

  The equation z3 - i. 4 sqrt(3) = 4 has a solution z1 = 2(cos(pi/9) + i sin(pi/9)) Find the other roots z2 and z3.

  
  

  De equation has the form z³=c. The three roots are in the Gauss-plane the vertices of an equilateral triangle. The three roots have 2 as modulus and the arguments of z₂ and z₃ arise by increasing the argument of z₁ by 2pi/3 and 4pi/3.

   
    z2 = 2( cos(pi/9 + 2 pi/3) + i sin(pi/9 + 2 pi/3)) =  2(cos(7 pi/9) + i sin(7 pi/9))
   
  
    z3 = 2( cos(pi/9 + 4 pi/3) + i sin(pi/9 + 4 pi/3)) =  2(cos(13 pi/9) + i sin(13 pi/9))
   
  

• 
  

  The number u, different from 1, is a solution of z3=1. Find the value of the determinant D = |1 u u2| |u u2 1| |u2 u 1|

  
  

  Since u is a root of z³-1= 0 and u is not 1, we have
  u³-1=0 <=> (u-1)(u²+u+1)=0 <=> u²+u+1=0

  the determinant does not change value if we add a column to another column. So, we add the second and the third column to the first one. D =

   
     |1+u+u2   u u2|
     |u+u2+1   u2 1|
     |u2+u+1  u   1|
  

  D = 0 because the first column is a 0-column.