Themes > Science > Mathematics > Algebra > Foci of a conic section > Topics and Problems > Problems about Complex numbers

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


READ THIS FIRST

If a problem is solved. It is not 'the' answer.
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


  •  
i.i.i.i=


     
1

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

  •  

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)+isqrt(2)
 
+sqrt(2)-isqrt(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 = - pi/6
(i-sqrt(3)) = 2.(cos(-pi/6) + i sin(-pi/6))
  • Simple calculations 

     
2.(cos(1) +isin(1)).5.(cos(2) +isin(2))= 10.(cos(3) +isin(3))

6.(cos(5) +isin(5))
--------------------= 2.(cos(3) +isin(3))
3.(cos(2) +isin(2))


(2.(cos(3) +isin(3)))5  =  32.(cos(15) +isin(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)+ isin(3)
                        _
Prove that 1 + z = (1 + z )z


     

     _           _
(1 + z )z = (z + z.z) = cos(3)+ isin(3) + (cos(3)+ isin(3))(cos(3) - isin(3))

= cos(3)+ isin(3) + 1 = 1 + z

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)+ isin(2)) =  2 cos2 (1) + i.2.sin(1).cos(1)

                      = 2.cos(1) (cos(1) + i sin(1))

(1 + cos(2)+ isin(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)+isin(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 (nt) + 2 i sin(n t) cos(n t)

                = 2 cos(nt). (cos(t) + i sin(t))


            zn            1
         --------  = ----------------  and this is real
         1 + z2n     2 cos(nt)
  • 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, zn has modulus 2n and argument n.pi/3.
    zn is real if and only if the argument is k.pi (with k = integer).
    So, zn 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 bigger 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)3 > 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

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


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