Equation: e^(ix) + 1 = 0
Working Out:
Rewrite e^(ix) in terms of trigonometric functions using Euler's formula:
cos(x) + i*sin(x) + 1 = 0
Now, separate the real and imaginary parts of the equation:
cos(x) + 1 = 0 and sin(x) = 0
Solve for x in both equations:
x = π and x = 2πk (where k is an integer)
2. Equation: |z + 3i| = |4z - 2|
Working Out:
For the absolute values to be equal, the expressions inside them must be equal or opposite:
z + 3i = 4z - 2 or z + 3i = -(4z - 2)
Solve for z in both cases:
z - 4z = -2 - 3i or z + 4z = -2 - 3i
-3z = -2 - 3i or 5z = -2 - 3i
Divide both sides by -3 or 5 to find the value of z:
z = (2 + 3i) / 3 or z = -(2 + 3i) / 5
3. Equation: (a + bi)^3 = 27 - 9i
Working Out:
Expand the left side using the cube of a binomial formula:
(a + bi)^3 = a^3 + 3a^2bi + 3ab^2i^2 + b^3i^3
(a + bi)^3 = a^3 + 3a^2bi - 3ab^2 - b^3i
Now set the equation equal to 27 - 9i:
a^3 + 3a^2bi - 3ab^2 - b^3i = 27 - 9i
Rearrange the equation to standard form (a + bi) and set the real and imaginary parts equal to each other:
a^3 - 3ab^2 = 27 and 3a^2b - b^3 = -9
Solve for a and b in both equations.
4. Equation: log(x + 3i) = log(5x - 2)
Working Out:
Using the properties of logarithms, the equation can be rewritten as:
x + 3i = 5x - 2
Working Out:
Rewrite e^(ix) in terms of trigonometric functions using Euler's formula:
cos(x) + i*sin(x) + 1 = 0
Now, separate the real and imaginary parts of the equation:
cos(x) + 1 = 0 and sin(x) = 0
Solve for x in both equations:
x = π and x = 2πk (where k is an integer)
2. Equation: |z + 3i| = |4z - 2|
Working Out:
For the absolute values to be equal, the expressions inside them must be equal or opposite:
z + 3i = 4z - 2 or z + 3i = -(4z - 2)
Solve for z in both cases:
z - 4z = -2 - 3i or z + 4z = -2 - 3i
-3z = -2 - 3i or 5z = -2 - 3i
Divide both sides by -3 or 5 to find the value of z:
z = (2 + 3i) / 3 or z = -(2 + 3i) / 5
3. Equation: (a + bi)^3 = 27 - 9i
Working Out:
Expand the left side using the cube of a binomial formula:
(a + bi)^3 = a^3 + 3a^2bi + 3ab^2i^2 + b^3i^3
(a + bi)^3 = a^3 + 3a^2bi - 3ab^2 - b^3i
Now set the equation equal to 27 - 9i:
a^3 + 3a^2bi - 3ab^2 - b^3i = 27 - 9i
Rearrange the equation to standard form (a + bi) and set the real and imaginary parts equal to each other:
a^3 - 3ab^2 = 27 and 3a^2b - b^3 = -9
Solve for a and b in both equations.
4. Equation: log(x + 3i) = log(5x - 2)
Working Out:
Using the properties of logarithms, the equation can be rewritten as:
x + 3i = 5x - 2
