(a) Compute the Modulus
The modulus of a complex number \( u = a + bi \) is given by:
\(
|u| = \sqrt{a^2 + b^2}
\)
For \( u = -1 + \sqrt{3} i \):
\(
|u| = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2
\)
Compute the Argument
The argument of \( u \), denoted as \( \arg(u) \), is given by:
\(
\theta = \tan^{-1} \left( \frac{b}{a} \right)
\)
Substituting \( a = -1 \) and \( b = \sqrt{3} \):
\(
\theta = \tan^{-1} \left( \frac{\sqrt{3}}{-1} \right) = \tan^{-1} (-\sqrt{3})
\)
Since the complex number lies in the **second quadrant**, we use:
\(
\theta = \pi – \frac{\pi}{3} = \frac{2\pi}{3}
\)
Express in Polar Form
The polar form of a complex number is:
\(
u = r e^{i\theta}
\)
Thus,
\(
u = 2 e^{i \frac{2\pi}{3}}
\)
(b)(i) Condition for \( u^n \) to be Real
A complex number is real if its imaginary part is zero. Using De Moivre’s Theorem:
\(
u^n = 2^n e^{i \frac{2n\pi}{3}}
\)
For this to be real, the exponent \( \frac{2n\pi}{3} \) must be an integer multiple of \( \pi \):
\(
\frac{2n\pi}{3} = k\pi, \quad k \in \mathbb{Z}
\)
Canceling \( \pi \) on both sides:
\(
\frac{2n}{3} = k
\)
Solve for Smallest Positive \( n \)
Rearrange for \( n \):
\(
2n = 3k
\)
\(
n = \frac{3k}{2}
\)
For \( n \) to be an integer, \( k \) must be even. The smallest positive \( k \) is 2:
\(
n = \frac{3(2)}{2} = 3
\)
Thus, the smallest \( n \) such that \( u^n \) is real is \( n = 3 \).
(b)(ii) Compute \( u^3 \)
Using De Moivre’s Theorem:
\(
u^3 = (2 e^{i \frac{2\pi}{3}})^3 = 2^3 e^{i \frac{6\pi}{3}}
\)
\(
= 8 e^{i 2\pi}
\)
Since \( e^{i 2\pi} = 1 \):
\(
u^3 = 8
\)
Thus, the value of \( u^3 \) is 8.
(c)(i) Identify a Given Root
We are given that \( u = -1 + \sqrt{3} i \) is a root. By the conjugate root theorem, since the coefficients of the polynomial are real, its complex conjugate is also a root:
\(
-1 – \sqrt{3} i
\)
Find the Third Root
Let the third root be \( c \). The sum of the roots is given by Vieta’s formulas:
\(
(-1+\sqrt{3}i) + (-1-\sqrt{3}i) + c = -\frac{5}{1}
\)
\(
-2 + c = -5
\)
Solving for \( c \):
\(
c = -3
\)
Thus, the third root is \( -3 \).
(c)(ii) Solve the Equation \( 1 + 5w + 10w^2 + 12w^3 = 0 \)**
Using the transformation \( w = \frac{1}{z} \), we substitute \( z = 1/w \):
\(
1 + 5w + 10w^2 + 12w^3 = 0
\)
Comparing with \( z^3 + 5z^2 + 10z + 12 = 0 \), we recognize that the roots of this equation are:
\(
w = \frac{1}{z}
\)
Thus, the roots are:
\(
w = \frac{1}{-1+\sqrt{3}i}, \quad w = \frac{1}{-1-\sqrt{3}i}, \quad w = \frac{1}{-3}
\)
Using the conjugate method, multiply numerator and denominator by the conjugate:
\(
w = \frac{1}{-1+\sqrt{3}i} \times \frac{-1-\sqrt{3}i}{-1-\sqrt{3}i}
\)
\(
= \frac{-1-\sqrt{3}i}{1 – (-3)} = \frac{-1-\sqrt{3}i}{4}
\)
Similarly,
\(
w = \frac{-1+\sqrt{3}i}{4}, \quad w = -\frac{1}{3}
\)
Thus, the roots are:
\(
w = \frac{-1 \pm \sqrt{3} i}{4}, \quad w = -\frac{1}{3}
\)
(d) Given:
\(
(a + bi)^2 = 2(a – bi)
\)
Expanding both sides:
\(
a^2 – b^2 + 2abi = 2a – 2bi
\)
Equating real and imaginary parts:
1. Real part equation:
\(
a^2 – b^2 = 2a
\)
2. Imaginary part equation:
\(
2ab = -2b
\)
Factor \( 2b \):
\(
2b(a+1) = 0
\)
So either \( b = 0 \) or \( a + 1 = 0 \). If \( b = 0 \), then solving \( a^2 = 2a \) gives \( a = 0 \) or \( a = 2 \), which is the real root.
If \( a = -1 \), solving \( a^2 – b^2 = 2a \):
\(
(-1)^2 – b^2 = 2(-1)
\)
\(
1 – b^2 = -2
\)
\(
b^2 = 3
\)
\(
b = \pm \sqrt{3}
\)
Thus, the solutions are:
\(
z = 2 \quad \text{(real root)}, \quad z = -1 \pm \sqrt{3} i \quad \text{(complex roots)}
\)