-
(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+2i-1
------- = i
1+1
-
1
So
4n
i = 1 (if n is a positive integer)
-
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))
-
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
-
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
-
( x > 2 and y = 0 )
-
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
-
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
-
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