I'm starting to read a book on complex analysis, and I'm having some troubles envolving simple equations with complex numbers. How can I solve equations envolving these numbers- what methods and strategies do you recommend? When it envolves aspects like $\bar z$, $|z|$ and $Arg(z)$, what should I do? Please take this equation as an example:$$|z|-z=1+2i$$
[Math] Solving simple complex equations
complex numberscomplex-analysis
Related Solutions
The first task is to clear the fractions, making it easier to collect like terms. Note that dividing by $-j$ amounts to multiplying by $j$, and that integer denominators can be cleared by multiplying by their least common multiple. Thus:
$$ \frac{V_a - V_s}{-15j} + \frac{V_a}{33} + \frac{V_a - V_o}{-25j} = 0 $$
becomes, after multiplying by $33\cdot 25$:
$$ (55j)(V_a - V_s) + 25V_a + (33j)(V_a - V_o) = 0 $$
After collecting like terms:
$$ (25+88j)V_a - 33j\; V_o - 55j\; V_s = 0 \tag{1} $$
Similar treatment of the second equation:
$$ \frac{V_o - V_a}{-25j} + \frac{V_o - V_s}{10} = 0 $$
results in:
$$ (2j)(V_o - V_a) + 5(V_o - V_s) = 0 $$
and then:
$$ -2j\; V_a + (5+2j)V_o - 5V_s = 0 \tag{2} $$
A moment's study reveals that (1),(2) form an independent pair of homogeneous linear equations in the three unknowns $V_a,V_o,V_s$. Thus solutions will collectively be all scalar multiples of some particular solution, i.e. a one parameter family of solutions.
One approach would be to set (for example) $V_s = 1$ and solve for the other two values. This yields a linear system of two equations in two unknowns:
$$ (25+88j)V_a - 33j\; V_o = 55j \tag{1'} $$
$$ -2j\; V_a + (5+2j)V_o = 5 \tag{2'} $$
The augmented matrix for this system is:
$$ \begin{bmatrix} 25+88j & -33j & \mid & 55j \\ -2j & 5+2j & \mid & 5 \end{bmatrix} $$
Putting the augmented matrix in reduced row echelon form is much the same use of elementary row operations, modified only insofar as complex arithmetic replaces real arithmetic.
Since it is desirable to do this "by hand", we will introduce a leading one in the first row by adding $44 - 6j$ times the second row to the first row (rather than by dividing the first row by $25 + 88j$):
$$ \begin{bmatrix} 1 & 244-5j & \mid & 220-5j \\ -2j & 5+2j & \mid & 5 \end{bmatrix} $$
Now I would add $2j$ times the first row to the second, to eliminate the entry under the new leading one:
$$ \begin{bmatrix} 1 & 244-5j & \mid & 220-5j \\ 0 & 15+490j & \mid & 15+440j \end{bmatrix} $$
Unfortunately we cannot put off doing a complex division much longer. To get a leading one in the second, I would divide by the coefficient $15+490j$:
$$ \frac{15+440j}{15+490j} = \frac{3+88j}{3+98j} = 1 - \frac{10j}{3+98j} $$
The denominator of this last fraction can be rationalized:
$$ \frac{10j}{3+98j} = \frac{10j(3-98j)}{3^2 + 98^2} = \frac{980+30j}{9613} $$
The (unreduced) row echelon form of the matrix is:
$$ \begin{bmatrix} 1 & 244-5j & \mid & 220-5j \\ 0 & 1 & \mid & 1 - \frac{980+30j}{9613} \end{bmatrix} $$
This can happen in several ways - indeed, in several senses:
- For any complex number $z$, $z$ and its complex conjugate $z^\ast$ have real sum and non-negative real product. This is relevant in the roots of polynomials with real coefficients, including the characteristic polynomials of $2\times 2$ matrices (whose eigenvalues have sum equal to its trace and product equal to its determinant).
- Complex inner products satisfy $\langle x|x\rangle\ge 0$, so naturally $\langle x+y|x+y\rangle-\langle x|x\rangle-\langle y|y\rangle$ is real. But it's $\langle x|y\rangle+\langle y|x\rangle$. Luckily, this is a sum of complex conjugates.
- Complex inner products are important in quantum mechanics because their square moduli are probabilities. These are of the form $zz^\ast$, so as aforesaid are non-negative.
- Quantum-mechanical observables have real eigenvalues. They're (typically infinite-dimensional) complex-valued matrices, but the complexification "cancels" in an unusual way: the identity $X_{ij}=X_{ji}^\ast$ ensures real eigenvalues.
- Raising an imaginary number to an even power makes it real. This arises in such physics as the Klein-Gordon equation.
- Cardano discovered how to solve cubic equations. Even when all roots are real, complex numbers are needed in the technique, but their imaginary parts cancel. The need for complex numbers is called the casus irreducibilis. This was the main historical impetus for taking complex numbers seriously (mathematicians didn't mind considering some quadratics insoluble, as opposed to having non-real roots). A more complicated version of this phenomenon is also found if we solve quartics. Cubics aren't that common in physics, but you'll find them in e.g. the climate of oceans.
Best Answer
For your specific question, remember that $z$ can be written as $z=x+iy$ and $|z|=\sqrt{x^2+y^2}$ is a real number. Also, $a+bi=x+yi$ if and only if $a=x$ and $b=y$ at the same time.
There are some handy tricks when dealing with complex number equations. Many of them will involve replacing $z$ with one of these:
where $\theta=Arg(z)$. Then you will be able to apply some formulas and identities and to get your result.
Sometimes it may be useful to think about it as a point in the complex plane. Then the trigonometric representation will make more sense.
Other times, when dealing with more analytical expressions (such as complex polynomials), it may be better to think in terms of the exponential.
Sometimes it will suffice to compare the real and imaginary terms (as in your exercise).
In the end, it all comes down to practice and familiarization. Do many exercises, work out some proofs in your books and try to work with them in other subjects, like Linear Algebra. Once you do, you'll have one of the most powerful tools of mathematics and physics at your disposal.