أقوى 50 تمريناً ومسألة في الأعداد العقدية لكل المستويات مع الحلول المفصلة

Exercise: z1 = 3 + 2i , z2 = 1 – 4i. Find z = z1 + z2

Solution: z = (3 + 1) + (2 – 4)i = 4 – 2i

Exercise: z1 = 5 – i , z2 = 2 + 3i. Find z = z1 – z2

Solution: z = (5 – 2) + (-1 – 3)i = 3 – 4i

Exercise: z1 = 2 + i , z2 = 3 – 2i. Find z = z1 * z2

Solution: z = (2)(3) – (2)(2i) + (i)(3) – (i)(2i) = 6 – 4i + 3i – 2(i^2) = 6 – i – 2(-1) = 8 – i

Exercise: z = (4 + 2i) / (1 – i)

Solution: z = [(4 + 2i)(1 + i)] / [(1 – i)(1 + i)] = (4 + 4i + 2i – 2) / (1 + 1) = (2 + 6i) / 2 = 1 + 3i

Exercise: z = 3 – 4i. Find |z|

Solution: |z| = sqrt(3^2 + (-4)^2) = sqrt(9 + 16) = sqrt(25) = 5

Exercise: z = -7 + 5i. Find z_bar

Solution: z_bar = -7 – 5i

Exercise: Find i^23

Solution: i^23 = i^(4*5 + 3) = (i^4)^5 * i^3 = 1^5 * (-i) = -i

Exercise: z = (1 + i)^2

Solution: z = 1^2 + 2(1)(i) + i^2 = 1 + 2i – 1 = 2i

Exercise: 2z – 3i = 4 + i

Solution: 2z = 4 + 4i => z = 2 + 2i

Exercise: z = 5e^(i * pi/2)

Solution: z = 5(cos(pi/2) + i*sin(pi/2)) = 5(0 + i*1) = 5i. Re(z) = 0, Im(z) = 5

Exercise: z = 1 – i

Solution: |z| = sqrt(2). arg(z) = -pi/4. z = sqrt(2)(cos(-pi/4) + i*sin(-pi/4))

Exercise: z = -1 + i*sqrt(3)

Solution: |z| = 2. arg(z) = 2pi/3. z = 2e^(i * 2pi/3)

Exercise: z^2 = -9

Solution: z = sqrt(-9) = +/- 3i

Exercise: Prove that |z1 * z2| = |z1| * |z2|

Solution: Let z1=r1*e^(i*a), z2=r2*e^(i*b). z1*z2 = r1*r2*e^(i*(a+b)). Modulus is r1*r2 = |z1|*|z2|. Proved.

Exercise: z^2 – 4z + 5 = 0

Solution: Delta = 16 – 20 = -4. Roots: z = (4 +/- 2i)/2 => z1 = 2 + i, z2 = 2 – i

Exercise: Calculate (sqrt(3) + i)^6

Solution: z = 2e^(i*pi/6). z^6 = 2^6 * e^(i*pi) = 64 * (-1) = -64

Exercise: Find roots of z^3 = 1

Solution: z_k = e^(i * 2k*pi/3) for k=0,1,2. z0 = 1, z1 = -1/2 + i*sqrt(3)/2, z2 = -1/2 – i*sqrt(3)/2

Exercise: |z – 2| = 3. What is the geometric locus?

Solution: Let z = x + iy. |(x-2) + iy| = 3 => (x-2)^2 + y^2 = 9. It is a circle centered at (2,0) with radius R=3.

Exercise: |z – i| = |z + 1|

Solution: Distance from point to (0,1) equals distance to (-1,0). Locus is the perpendicular bisector of the segment connecting these points. Equation: y = -x.

Exercise: Simplify z = (1+i)^10 / (1-i)^8

Solution: (1+i)^2 = 2i => (1+i)^10 = (2i)^5 = 32i. (1-i)^2 = -2i => (1-i)^8 = (-2i)^4 = 16. z = 32i / 16 = 2i.

Exercise: z1 + z2 = 2 ; z1 – z2 = 2i

Solution: Add them: 2z1 = 2 + 2i => z1 = 1 + i. Subtract: 2z2 = 2 – 2i => z2 = 1 – i.

Exercise: w^2 = 3 + 4i

Solution: Let w = x + iy. x^2 – y^2 = 3, 2xy = 4, x^2 + y^2 = 5. Adding 1st and 3rd: 2x^2 = 8 => x = +/- 2. y = 2/x. w1 = 2 + i, w2 = -2 – i.

Exercise: Sum S = 1 + cos(x) + cos(2x). Express using Euler’s.

Solution: S = Re(1 + e^(ix) + e^(i2x)). It’s a geometric series Re((1-e^(i3x))/(1-e^(ix))).

Exercise: z’ = z + 3 – 2i

Solution: This is a translation (انسحاب) by the vector V(3, -2).

Exercise: z’ = i * z

Solution: Multiply by i = e^(i*pi/2). This is a rotation (دوران) around the origin by angle +pi/2.

Exercise: Center C(1+i), passes through A(4+5i). Write the locus.

Solution: Radius R = |z_A – z_C| = |3 + 4i| = 5. Locus equation: |z – (1+i)| = 5.

Exercise: Points A(i), B(2), C(3+2i). Prove triangle ABC is right at A.

Solution: (z_C – z_A) / (z_B – z_A) = (3+i) / (2-i) = (3+i)(2+i)/5 = (5+5i)/5 = 1+i. The argument is pi/4. Wait, let’s re-verify. Angle at B: (z_C-z_B)/(z_A-z_B) = (1+2i)/(-2+i) = -i. Argument is -pi/2. So it is right-angled at B, not A!

Exercise: Evaluate E = (cos(pi/5) + i*sin(pi/5))^10

Solution: By De Moivre’s: E = cos(10*pi/5) + i*sin(10*pi/5) = cos(2pi) + i*sin(2pi) = 1 + 0i = 1.

Exercise: z^3 – 8 = 0

Solution: z^3 = 8. z_k = 2 * e^(i * 2k*pi/3). z0=2, z1=-1+i*sqrt(3), z2=-1-i*sqrt(3).

Exercise: Prove z = (2+i)^4 + (2-i)^4 is a real number.

Solution: Let w = (2+i)^4. Then z = w + w_bar. Since any number plus its conjugate equals 2*Re(w), the sum is strictly real. Im(z) = 0.

Exercise: Solve in C: z^4 – (1+2i)z^2 – (1-i) = 0

Solution:
Let u = z^2. The equation becomes a quadratic in u:
u^2 – (1+2i)u – (1-i) = 0
Calculate the discriminant Delta:
Delta = (1+2i)^2 – 4(-1+i)
Delta = (1 + 4i – 4) + 4 – 4i = -3 + 4 = 1
Roots for u:
u1 = [(1+2i) + 1]/2 = 1 + i
u2 = [(1+2i) – 1]/2 = i
Now solve z^2 = u1 and z^2 = u2:
1) z^2 = i => z = +/- (sqrt(2)/2 + i*sqrt(2)/2)
2) z^2 = 1+i => r = sqrt(2), theta = pi/4. z = +/- 2^(1/4) * e^(i*pi/8).
Four complex roots in total.

Exercise: Find the locus of point M(z) such that Z = (z – 2i) / (z + 1) is purely imaginary.

Solution:
Z is purely imaginary iff Re(Z) = 0 and z != -1.
Let z = x + iy. Z = (x + i(y-2)) / ((x+1) + iy).
Multiply numerator and denominator by the conjugate of denominator: ((x+1) – iy).
Numerator = [x(x+1) + y(y-2)] + i[…]
For Re(Z) = 0: x(x+1) + y(y-2) = 0
x^2 + x + y^2 – 2y = 0
Complete the square: (x + 1/2)^2 – 1/4 + (y – 1)^2 – 1 = 0
(x + 1/2)^2 + (y – 1)^2 = 5/4
The locus is a circle centered at C(-1/2, 1) with radius R = sqrt(5)/2, excluding the point (-1, 0).

Exercise: A transformation f maps z to z’ = (1+i)z + 2-i. Determine the nature, center, ratio, and angle of f.

Solution:
Form: z’ = az + b. Here a = 1+i, b = 2-i.
Since a != 1, it’s a direct similitude (تشابه مباشر).
Ratio k = |a| = |1+i| = sqrt(2).
Angle theta = arg(a) = pi/4.
Center omega is the invariant point: w = aw + b => w = b / (1 – a).
w = (2-i) / (1 – (1+i)) = (2-i) / (-i) = (2-i)(i) / 1 = 2i + 1.
Center is Omega(1, 2).

Exercise: Let z_0 = 1, z_{n+1} = (1/2 + i*sqrt(3)/2) * z_n. Find z_n as a function of n and determine if it converges.

Solution:
The factor a = 1/2 + i*sqrt(3)/2 = e^(i*pi/3).
This is a geometric sequence: z_n = z_0 * a^n = (e^(i*pi/3))^n = e^(i * n*pi/3).
|z_n| = 1 for all n. The sequence of points lies on the unit circle.
It does not converge to a single limit, it is periodic with period 6, since a^6 = e^(2pi*i) = 1.

Exercise: Solve cos(3x) + cos(x) = 0 using complex exponentials.

Solution:
Let z = e^(ix). cos(x) = (z + 1/z)/2, cos(3x) = (z^3 + 1/z^3)/2.
Equation: z^3 + 1/z^3 + z + 1/z = 0.
Multiply by z^3: z^6 + z^4 + z^2 + 1 = 0.
Factor: z^4(z^2 + 1) + 1(z^2 + 1) = 0 => (z^4 + 1)(z^2 + 1) = 0.
Case 1: z^2 = -1 => z = +/- i => e^(ix) = e^(i*pi/2) or e^(-i*pi/2) => x = pi/2 + k*pi.
Case 2: z^4 = -1 => z = e^(i*(pi+2k*pi)/4). x = pi/4, 3pi/4, 5pi/4, 7pi/4 + 2k*pi.
General solution combines these angles.

Exercise: Find the roots of (z – i)^5 = (z + i)^5.

Solution:
Clearly z != -i. Divide: ((z-i)/(z+i))^5 = 1.
Let w = (z-i)/(z+i). w_k = e^(i*2k*pi/5) for k=0,1,2,3,4.
z – i = w_k(z + i) => z(1 – w_k) = i(1 + w_k) => z_k = i(1 + w_k)/(1 – w_k).
Using w_k = cos(A) + i*sin(A) where A = 2k*pi/5:
z_k = i * [2*cos(A/2)*e^(i*A/2)] / [-2i*sin(A/2)*e^(i*A/2)]
z_k = -cot(k*pi/5).
For k=0, undefined (implies degree 4 equation practically). Roots are strictly real.

Exercise: Let j = e^(i*2pi/3). Prove that (a + b*j + c*j^2)(a + b*j^2 + c*j) = a^2 + b^2 + c^2 – ab – bc – ca.

Solution:
Note that 1 + j + j^2 = 0 and j^3 = 1. Conjugate of j is j^2.
Expand the product:
= a^2 + ab*j^2 + ac*j + ab*j + b^2*j^3 + bc*j^2 + ac*j^2 + bc*j^4 + c^2*j^3
Substitute j^3 = 1 and j^4 = j:
= a^2 + b^2 + c^2 + ab(j + j^2) + bc(j^2 + j) + ac(j + j^2).
Since j + j^2 = -1:
= a^2 + b^2 + c^2 – ab – bc – ac. Proved.

Exercise: Solve z^3 – 3iz^2 – 3z + i = 0

Solution:
Notice the binomial expansion structure.
(z – i)^3 = z^3 – 3z^2(i) + 3z(i^2) – i^3
= z^3 – 3iz^2 – 3z + i.
Thus, the equation is exactly (z – i)^3 = 0.
Solution: z = i (a root with multiplicity 3).

Exercise: Compute Sum(k=0 to n) of C(n, k)*cos(k*x).

Solution:
Let S = Sum C(n,k) cos(kx) and T = Sum C(n,k) sin(kx).
S + iT = Sum C(n,k) e^(ikx) = Sum C(n,k) (e^(ix))^k * 1^(n-k).
By Newton’s Binomial Theorem: S + iT = (1 + e^(ix))^n.
1 + e^(ix) = e^(ix/2)(e^(-ix/2) + e^(ix/2)) = 2*cos(x/2)*e^(ix/2).
(1 + e^(ix))^n = 2^n * cos^n(x/2) * e^(in*x/2).
Taking the real part:
S = 2^n * cos^n(x/2) * cos(nx/2).

Exercise: Find the image of the upper half-plane Im(z) > 0 under the mapping w = (z – i)/(z + i).

Solution:
Let z = x + iy, where y > 0.
|w| = |x + i(y-1)| / |x + i(y+1)|
|w|^2 = [x^2 + (y-1)^2] / [x^2 + (y+1)^2] Since y > 0, (y-1)^2 < (y+1)^2. Therefore, x^2 + (y-1)^2 < x^2 + (y+1)^2.
This means |w|^2 < 1, so |w| < 1.
The image is the interior of the unit disk (قرص الوحدة) |w| < 1.

Exercise: Express cos(5x) as a polynomial in cos(x).

Solution:
(cos(x) + i*sin(x))^5 = cos(5x) + i*sin(5x).
Expand LHS using Binomial theorem, then equate the real parts:
cos(5x) = cos^5(x) – C(5,2)*cos^3(x)*sin^2(x) + C(5,4)*cos(x)*sin^4(x)
sin^2(x) = 1 – cos^2(x), sin^4(x) = (1 – cos^2(x))^2 = 1 – 2cos^2(x) + cos^4(x).
cos(5x) = cos^5(x) – 10*cos^3(x)(1 – cos^2(x)) + 5*cos(x)(1 – 2cos^2(x) + cos^4(x))
cos(5x) = cos^5(x) – 10*cos^3(x) + 10*cos^5(x) + 5*cos(x) – 10*cos^3(x) + 5*cos^5(x)
cos(5x) = 16*cos^5(x) – 20*cos^3(x) + 5*cos(x).

Exercise: Factor P(x) = x^4 + 1 into two quadratic polynomials with real coefficients.

Solution:
Find the complex roots: z^4 = -1 = e^(i*pi).
Roots are z_k = e^(i*(pi + 2k*pi)/4) for k=0,1,2,3.
z0 = e^(i*pi/4), z1 = e^(i*3pi/4), z2 = e^(i*5pi/4) = z1_bar, z3 = e^(i*7pi/4) = z0_bar.
Group conjugate roots to form real quadratics:
Q1(x) = (x – z0)(x – z0_bar) = x^2 – 2x*Re(z0) + |z0|^2 = x^2 – sqrt(2)x + 1.
Q2(x) = (x – z1)(x – z1_bar) = x^2 – 2x*Re(z1) + |z1|^2 = x^2 + sqrt(2)x + 1.
P(x) = (x^2 – sqrt(2)x + 1)(x^2 + sqrt(2)x + 1).

Exercise: Let A(z_a), B(z_b), C(z_c), D(z_d). Prove that lines (AB) and (CD) are orthogonal iff Re((z_b – z_a)/(z_d – z_c)) = 0.

Solution:
Let W = (z_b – z_a)/(z_d – z_c). The argument of W represents the angle between vectors CD and AB.
arg(W) = arg(z_b – z_a) – arg(z_d – z_c) = angle(CD, AB).
(AB) is orthogonal to (CD) iff angle(CD, AB) = +/- pi/2 [mod pi].
This is true iff W is purely imaginary.
W is purely imaginary iff Re(W) = 0. Proved.

Exercise: Solve the equation: z^2 + 2*z_bar = 0.

Solution:
Let z = r*e^(i*theta). z^2 = r^2*e^(2i*theta), z_bar = r*e^(-i*theta).
r^2*e^(2i*theta) = -2*r*e^(-i*theta).
If r = 0, z = 0 is a solution.
If r != 0, divide by r: r*e^(2i*theta) = 2*e^(i*pi)*e^(-i*theta) = 2*e^(i*(pi – theta)).
Equate moduli: r = 2.
Equate arguments: 2*theta = pi – theta + 2k*pi => 3*theta = pi + 2k*pi => theta = pi/3 + 2k*pi/3.
k=0: theta = pi/3 => z1 = 2(1/2 + i*sqrt(3)/2) = 1 + i*sqrt(3).
k=1: theta = pi => z2 = -2.
k=2: theta = 5pi/3 => z3 = 1 – i*sqrt(3).
Solutions: {0, -2, 1+i*sqrt(3), 1-i*sqrt(3)}.

Exercise: Find the invariant points of f(z) = (iz – 1)/(z + i).

Solution:
Set f(z) = z => (iz – 1)/(z + i) = z.
iz – 1 = z(z + i) => iz – 1 = z^2 + iz.
Subtract iz from both sides: z^2 = -1.
Thus z = i or z = -i.
However, z = -i is a singularity of f(z) (makes denominator zero), so it is rejected in C.
The only invariant point is z = i.

Exercise: For any complex numbers z1, z2, z3, z4, prove |z1-z3||z2-z4| <= |z1-z2||z3-z4| + |z1-z4||z2-z3|.

Solution:
This is Ptolemy’s Inequality.
Consider the algebraic identity: (z1-z3)(z2-z4) = (z1-z2)(z3-z4) + (z1-z4)(z2-z3).
Proof of identity: Expand RHS: z1z3 – z1z4 – z2z3 + z2z4 + z1z2 – z1z3 – z4z2 + z4z3.
Wait, standard algebraic identity is: (a-b)(c-d) + (a-d)(b-c) = (a-c)(b-d). Let’s check.
ac – ad – bc + bd + ab – ac – db + dc = ab – ad – bc + dc. (Matches (a-d)(b-c) logic).
Taking the modulus of both sides and applying the triangle inequality |A + B| <= |A| + |B|:
|(z1-z3)(z2-z4)| = |(z1-z2)(z3-z4) + (z1-z4)(z2-z3)|
|z1-z3||z2-z4| <= |z1-z2||z3-z4| + |z1-z4||z2-z3|. Proved.

Exercise: Find locus of z such that W = (z-1)/(z+1) is a strictly positive real number.

Solution:
W in R+* implies arg(W) = 0 [2pi].
arg((z-1)/(z+1)) = arg(z-1) – arg(z+1) = angle(M(z), A(1), B(-1)).
The angle between vector MB and MA is 0.
This means points M, A, B are collinear, and M is outside the segment [AB].
Locus is the real axis excluding the interval [-1, 1]. In coordinates: y=0, and x in (-infinity, -1) U (1, +infinity).

Exercise: Solve in C: z^5 = 1, and use the roots to calculate cos(2pi/5).

Solution:
Roots are 1, w, w^2, w^3, w^4 where w = e^(i*2pi/5).
Sum of roots = 1 + w + w^2 + w^3 + w^4 = 0.
Note that w^4 = w_bar, and w^3 = (w^2)_bar.
Equation: 1 + 2Re(w) + 2Re(w^2) = 0.
1 + 2*cos(2pi/5) + 2*cos(4pi/5) = 0.
Use cos(4pi/5) = 2*cos^2(2pi/5) – 1.
Let x = cos(2pi/5). 1 + 2x + 2(2x^2 – 1) = 0 => 4x^2 + 2x – 1 = 0.
Delta = 4 + 16 = 20. x = (-2 + sqrt(20))/8 = (-1 + sqrt(5))/4. (Positive since 2pi/5 < pi/2).
cos(2pi/5) = (sqrt(5) – 1)/4.

Exercise: Solve the system: z1 + z2 = 4 , z1^3 + z2^3 = 52.

Solution:
Recall z1^3 + z2^3 = (z1 + z2)(z1^2 – z1*z2 + z2^2) = (z1 + z2)((z1 + z2)^2 – 3*z1*z2).
Substitute knowns: 52 = 4 * (16 – 3*z1*z2).
13 = 16 – 3*z1*z2 => 3*z1*z2 = 3 => z1*z2 = 1.
Now z1, z2 are roots of the quadratic equation X^2 – S*X + P = 0.
X^2 – 4X + 1 = 0.
Delta = 16 – 4 = 12 = (2*sqrt(3))^2.
Roots: X = (4 +/- 2*sqrt(3))/2 = 2 +/- sqrt(3).
The solutions are the pairs (2+sqrt(3), 2-sqrt(3)) and (2-sqrt(3), 2+sqrt(3)).

Exercise: Evaluate the line integral I = Integral(C) (z_bar) dz, where C is the unit circle |z| = 1, counter-clockwise.

Solution:
Parameterize the circle C: z = e^(i*t), where t from 0 to 2pi.
Then dz = i*e^(i*t) dt.
The conjugate of z on the unit circle is z_bar = e^(-i*t).
Substitute into the integral:
I = Integral(0 to 2pi) [e^(-i*t)] * [i*e^(i*t)] dt
I = Integral(0 to 2pi) i * e^(0) dt
I = i * Integral(0 to 2pi) 1 dt
I = i * [t] from 0 to 2pi
I = 2*pi*i.
Notice that this cannot be evaluated using Cauchy’s theorem directly because f(z) = z_bar is not analytic.