3.3 Potenzen und Wurzeln
Aus Online Mathematik Brückenkurs 2
| Theory | Exercises |
Contents:
- De Moivre's formula
- Binomial equations
- Exponential function
- Euler's formula
- Completing the square
- Quadratic equations
Learning outcomes:
After this section, you will have learned how to:
- Calculate the powers of complex numbers with de Moivre's formula.
- Calculate the roots of certain complex numbers by rewriting to polar form.
- Solve binomial equations.
- Complete the square for complex quadratic expressions.
- Solve complex quadratic equations.
De Moivre's formula
The computational rules \displaystyle \ \arg (zw) = \arg z + \arg w\ and \displaystyle \ |\,zw\,| = |\,z\,|\cdot|\,w\,|\ mean that
| \displaystyle \biggl\{\begin{align*}&\arg (z\cdot z) = \arg z + \arg z \\ &|\,z\cdot z\,| = |\,z\,|\cdot|\,z\,|\end{align*}\qquad\biggl\{\begin{align*}&\arg z^3 = 3 \arg z \cr &|\,z^3\,| = |\,z\,|^3\end{align*}\qquad\text{etc.} |
For an arbitrary number \displaystyle z=r\,(\cos \alpha +i\,\sin \alpha), we therefore have the following relationship
| \displaystyle z^n = \bigl(r\,(\cos \alpha +i\sin \alpha)\bigr)^n = r^n\,(\cos n\alpha +i\,\sin n\alpha)\,\mbox{.} |
If \displaystyle |\,z\,|=1, (i.e. \displaystyle z lies on the unit circle) then one has the special relationship
| \displaystyle (\cos \alpha +i\,\sin \alpha)^n = \cos n\alpha +i\,\sin n\alpha\,\mbox{,} |
which is usually referred to as de Moivres formula. This relationship is very useful when it comes to deriving trigonometric identities and calculating the roots and powers of complex numbers.
Example 1
If \displaystyle z = \frac{1+i}{\sqrt2}, determine \displaystyle z^3 and \displaystyle z^{100}.
We write \displaystyle z in polar form \displaystyle \ \ z= \frac{1}{\sqrt2} + \frac{i}{\sqrt2} = 1\cdot \Bigl(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\Bigr)\ \ and Moivre's formula gives
| \displaystyle \begin{align*}z^3 &= \Bigl( \cos\frac{\pi}{4} + i\,\sin\frac{\pi}{4}\,\Bigr)^3 = \cos\frac{3\pi}{4} + i\,\sin\frac{3\pi}{4} = -\frac{1}{\sqrt2} + \frac{1}{\sqrt2}\,i = \frac{-1+i}{\sqrt2}\,\mbox{,}\\[6pt] z^{100} &= \Bigl( \cos\frac{\pi}{4} + i\,\sin\frac{\pi}{4}\,\Bigr)^{100} = \cos\frac{100\pi}{4} + i\,\sin\frac{100\pi}{4}\\[4pt] &= \cos 25\pi + i\,\sin 25\pi = \cos \pi + i\,\sin \pi = -1\,\mbox{.}\end{align*} |
Example 2
In the usual way one does an expansion by means of the squaring rules
| \displaystyle \begin{align*} (\cos v + i\,\sin v)^2 &= \cos^2\!v + i^2 \sin^2\!v + 2i \sin v \cos v\\ &= \cos^2\!v - \sin^2\!v + 2i \sin v \cos v\end{align*} |
and according to de Moivre's formula one gets
| \displaystyle (\cos v + i \sin v)^2 = \cos 2v + i \sin 2v\,\mbox{.} |
If one equates the real and imaginary parts of the two expressions one gets the well-known trigonometric formulas
| \displaystyle \biggl\{\begin{align*}\cos 2v &= \cos^2\!v - \sin^2\!v\,\mbox{,}\\[2pt] \sin 2v&= 2 \sin v \cos v\,\mbox{.}\end{align*} |
Example 3
Simplify \displaystyle \ \ \frac{(\sqrt3 + i)^{14}}{(1+i\sqrt3\,)^7(1+i)^{10}}\,.
We write the numbers \displaystyle \sqrt{3}+i, \displaystyle 1+i\sqrt{3} and \displaystyle 1+i in polar form
- \displaystyle \quad\sqrt{3} + i = 2\Bigl(\cos\frac{\pi}{6} + i\,\sin\frac{\pi}{6}\,\Bigr)\vphantom{\biggl(},
- \displaystyle \quad 1+i\sqrt{3} = 2\Bigl(\cos\frac{\pi}{3} + i\,\sin\frac{\pi}{3}\,\Bigr)\vphantom{\biggl(},
- \displaystyle \quad 1+i = \sqrt2\,\Bigl(\cos\frac{\pi}{4} + i\,\sin\frac{\pi}{4}\,\Bigr)\vphantom{\biggl(}.
Then we get with de Moivre's formula
| \displaystyle \frac{(\sqrt3 + i)^{14}}{(1+i\sqrt3\,)^7(1+i)^{10}} = \frac{\displaystyle 2^{14}\Bigl(\cos\frac{14\pi}{6} + i\,\sin \frac{14\pi}{6}\,\Bigr)\vphantom{\biggl(}}{\displaystyle 2^7\Bigl(\cos \frac{7\pi}{3} + i\,\sin\frac{7\pi}{3}\,\Bigr) \cdot (\sqrt{2}\,)^{10}\Bigl(\cos\frac{10\pi}{4} + i\,\sin\frac{10\pi}{4}\,\Bigr)\vphantom{\biggl(}} |
and this expression can be simplified by performing multiplication and division in polar form
| \displaystyle \begin{align*}\frac{\displaystyle 2^{14}\Bigl(\cos\frac{14\pi}{6} + i\,\sin\frac{14\pi}{6}\,\Bigr)\vphantom{\biggl(}} {\displaystyle 2^{12}\Bigl(\cos\frac{29\pi}{6} + i\,\sin\frac{29\pi}{6}\,\Bigr)\vphantom{\biggl(}} &= 2^2 \Bigl(\cos\Bigl( -\frac{15\pi}{6}\,\Bigr) + i\,\sin\Bigl( -\frac{15\pi}{6}\,\Bigr)\,\Bigr)\\[8pt] &= 4\Bigl(\cos \Bigl( -\frac{\pi}{2}\,\Bigr) + i\,\sin\Bigl( -\frac{\pi}{2}\,\Bigr)\,\Bigr) = -4i\,\mbox{.}\end{align*} |
Binomial equations
A complex number \displaystyle z is called the nth root of the complex number \displaystyle w if
| \displaystyle z^n= w \mbox{.} |
The above relationship can also be seen as an equation in which \displaystyle z is unknown. This type of equation is called a binomial equation. The solutions are obtained by rewriting both sides in polar form and comparing both the moduli and the arguments.
For a given number \displaystyle w=|\,w\,|\,(\cos \theta + i\,\sin \theta) one assumes that \displaystyle z=r\,(\cos \alpha + i\, \sin \alpha) and after insertion, the binomial equation becomes
| \displaystyle r^{\,n}\,(\cos n\alpha + i \sin n\alpha) =|w|\,(\cos \theta + i \sin \theta)\,\mbox{,} |
where de Moivres formula has been used on the left-hand side. Equating moduli and arguments gives
| \displaystyle \biggl\{\begin{align*} r^{\,n} &= |w|\,\mbox{,}\\ n\alpha &= \theta + k\cdot 2\pi\,\mbox{.}\end{align*} |
Note that we add multiples of \displaystyle 2\pi to include all possible values of the argument that have the same direction as \displaystyle \theta. One gets
| \displaystyle \biggl\{\begin{align*} r &={\textstyle\sqrt[\scriptstyle n]{|w|}},\\ \alpha &= (\theta + 2k\pi)/n\,, \quad k=0, \pm 1, \pm 2, \ldots\end{align*} |
This gives one value of \displaystyle r, but infinitely many values of \displaystyle \alpha. Despite this, there are not infinitely many solutions. From \displaystyle k = 0 to \displaystyle k = n - 1 one gets different arguments for \displaystyle z and thus different positions for \displaystyle z in the complex plane. For the other values of \displaystyle k due to the periodicity of the sine and cosine, one returns to these positions and therefore no new solutions are obtained. This reasoning shows that the equation \displaystyle z^n=w has exactly \displaystyle n roots.
Comment. Note that the arguments of the roots differ from each other by \displaystyle 2\pi/n so that the roots are evenly distributed on a circle with radius \displaystyle \sqrt[\scriptstyle n]{|w|} and form corners in a regular n-gon (an n sided polygon).
Exempel 4
Solve the binomial equation \displaystyle \ z^4= 16\,i\,.
Write \displaystyle z and \displaystyle 16\,i in polar form
- \displaystyle \quad z=r\,(\cos \alpha + i\,\sin \alpha)\,,
- \displaystyle \quad 16\,i= 16\Bigl(\cos\frac{\pi}{2} + i\,\sin\frac{\pi}{2}\,\Bigr)\vphantom{\biggl(}.
This turns the equation \displaystyle \ z^4=16\,i\ into
| \displaystyle r^4\,(\cos 4\alpha + i\,\sin 4\alpha) = 16\Bigl(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\,\Bigr)\,\mbox{.} |
Matching the moduli and arguments on both sides gives
| \displaystyle \biggl\{\begin{align*} r^4 &= 16,\\ 4\alpha &= \pi/2 + k\cdot 2\pi,\end{align*}\qquad\text{i.e.}\qquad\biggl\{\begin{align*} r &= \sqrt[\scriptstyle 4]{16}= 2, \\ \alpha &= \pi/8 + k\pi/2\,,\quad k=0,1,2,3.\end{align*} |
|
The solutions to the equation are thus
| 3.3 - Figure - The complex numbers z₁, z₂, z₃ and z₄ |
Exponential form of complex numbers
If we manipulate \displaystyle i as if it were a real number and treat a complex number \displaystyle z as a function of just \displaystyle \alpha ( where \displaystyle r is a constant),
| \displaystyle f(\alpha) = r\,(\cos \alpha + i\,\sin \alpha) |
we get after differentiation
| \displaystyle \begin{align*} f^{\,\prime}(\alpha) &= -r\sin \alpha + r\,i\,\cos \alpha = r\,i^2 \sin \alpha + r\,i\,\cos \alpha = i\,r\,(\cos \alpha + i\,\sin \alpha) = i\,f(\alpha)\\ f^{\,\prime\prime} (\alpha) &= - r\,\cos \alpha - r\,i\,\sin \alpha = i^2\,r\,(\cos \alpha + i\,\sin \alpha) = i^2\, f(\alpha)\cr &\text{etc.}\end{align*} |
The only real-valued functions which behave like this are \displaystyle f(x)= e^{\,kx}, which justifies the definition
| \displaystyle e^{\,i\alpha} = \cos \alpha + i\,\sin \alpha\,\mbox{.} |
This definition turns out to be a completely natural generalisation of the exponential function for the real numbers. Putting \displaystyle z=a+ib one gets
| \displaystyle e^{\,z} = e^{\,a+ib} = e^{\,a} \cdot e^{\,ib} = e^{\,a}(\cos b + i\,\sin b)\,\mbox{.} |
The definition of \displaystyle e^{\,z} may be regarded as a convenient notation for the polar form of a complex number, as \displaystyle z=r\,(\cos \alpha + i\,\sin \alpha) = r\,e^{\,i\alpha}\,.
Example 5
For a real number \displaystyle z the definition is consistent with the case when the exponent is real, as \displaystyle z=a+0\cdot i which gives
| \displaystyle e^{\,z} = e^{\,a+0\cdot i} = e^a (\cos 0 + i \sin 0) = e^a \cdot 1 = e^a\,\mbox{.} |
Example 6
A further indication of why the above definition is so natural is given by the relationship
| \displaystyle \bigl(e^{\,i\alpha}\bigr)^n = (\cos \alpha + i \sin \alpha)^n = \cos n\alpha + i \sin n \alpha = e^{\,in\alpha}\,\mbox{,} |
which demonstrates that de Moivre's formula is actually identical to the well-known law of exponents,
| \displaystyle \left(a^x\right)^y = a^{x\,y}\,\mbox{.} |
Example 7
From the above definition, one can obtain the relationship
| \displaystyle e^{\pi\,i} = \cos \pi + i \sin \pi = -1 |
which connects together the, generally regarded, most basic numbers in mathematics: \displaystyle e, \displaystyle \pi, \displaystyle i and 1. This relationship is seen by many as the most beautiful in mathematics and was discovered by Euler in the early 1700's.
Example 8
Solve the equation \displaystyle \ (z+i)^3 = -8i.
Put \displaystyle w = z + i. We then get the binomial equation \displaystyle \ w^3=-8i\,. To begin with, we rewrite \displaystyle w and \displaystyle -8i in polar form
- \displaystyle \quad w=r\,(\cos \alpha + i\,\sin \alpha) = r\,e^{i\alpha}\,\mbox{,}
- \displaystyle \quad -8i = 8\Bigl(\cos \frac{3\pi}{2} + i\,\sin\frac{3\pi}{2}\,\Bigr) = 8\,e^{3\pi i/2}\vphantom{\biggl(}\,\mbox{.}
The equation in polar form is \displaystyle \ r^3e^{3\alpha i}=8\,e^{3\pi i/2}\ and matching the moduli and arguments on both sides gives,
| \displaystyle \biggl\{\begin{align*} r^3 &= 8\,\mbox{,}\\ 3\alpha &= 3\pi/2+2k\pi\,\mbox{,}\end{align*}\qquad\Leftrightarrow\qquad\biggl\{\begin{align*} r&=\sqrt[\scriptstyle 3]{8}\,\mbox{,}\\ \alpha&= \pi/2+2k\pi/3\,,\quad k=0,1,2\,\mbox{.}\end{align*} |
The roots of the equation are thus
- \displaystyle \quad w_1 = 2\,e^{\pi i/2} = 2\Bigl(\cos \frac{\pi}{2} + i\,\sin\frac{\pi}{2}\,\Bigr) = 2i\,\mbox{,}\quad\vphantom{\biggl(}
- \displaystyle \quad w_2 = 2\,e^{7\pi i/6} = 2\Bigl(\cos\frac{7\pi}{6} + i\,\sin\frac{7\pi}{6}\,\Bigr) = -\sqrt{3}-i\,\mbox{,}\quad\vphantom{\Biggl(}
- \displaystyle \quad w_3 = 2\,e^{11\pi i/6} = 2\Bigl(\cos\frac{11\pi}{6} + i\,\sin\frac{11\pi}{6}\,\Bigr) = \sqrt{3}-i\,\mbox{,}\quad\vphantom{\biggl(}
i.e. \displaystyle z_1 = 2i-i=i, \displaystyle z_2 = - \sqrt{3}-2i and \displaystyle z_3 = \sqrt{3}-2i.
Example 9
Solve \displaystyle \ z^2 = \overline{z}\,.
If for \displaystyle z=a+ib one has \displaystyle |\,z\,|=r and \displaystyle \arg z = \alpha then for \displaystyle \overline{z}= a-ib one gets \displaystyle |\,\overline{z}\,|=r and \displaystyle \arg \overline{z} = - \alpha.This means that \displaystyle z=r\,e^{i\alpha} and \displaystyle \overline{z} = r\,e^{-i\alpha}. The equation can be written
| \displaystyle (r\,e^{i\alpha})^2 = r\,e^{-i\alpha}\qquad\text{or}\qquad r^2 e^{2i\alpha}= r\,e^{-i\alpha}\,\mbox{,} |
which directly gives that \displaystyle r=0 is a solution, i.e. \displaystyle z=0. If we assume that \displaystyle r\not=0 then the equation can be written as \displaystyle \ r\,e^{3i\alpha} = 1\,, which gives after matching moduli and arguments
| \displaystyle \biggl\{\begin{align*} r &= 1\,\mbox{,}\\ 3\alpha &= 0 + 2k\pi\,\mbox{,}\end{align*}\qquad\Leftrightarrow\qquad\biggl\{\begin{align*} r &= 1\,\mbox{,}\\ \alpha &= 2k\pi/3\,\mbox{,}\quad k=0,1,2\,\mbox{.}\end{align*} |
The solutions are
- \displaystyle \quad z_1 = e^0 = 1\,\mbox{,}
- \displaystyle \quad z_2 = e^{2\pi i/ 3} = \cos\frac{2\pi}{3} + i\,\sin\frac{2\pi}{3} = -\frac{1}{2} + \frac{\sqrt3}{2}\,i\,\mbox{,}\vphantom{\Biggl(}
- \displaystyle \quad z_3 = e^{4\pi i/ 3} = \cos\frac{4\pi}{3} + i\,\sin\frac{4\pi}{3} = -\frac{1}{2} - \frac{\sqrt3}{2}\,i\,\mbox{,}
- \displaystyle \quad z_4 = 0\,\mbox{.}
Completing the square
The squaring rules,
| \displaystyle \left\{\begin{align*} (a+b)^2 &= a^2+2ab+b^2\\ (a-b)^2 &= a^2-2ab+b^2\end{align*}\right. |
which are usually used to expand parenthesis can also be used in reverse to obtain quadratic expressions. For example,
| \displaystyle \begin{align*} x^2+4x+4 &= (x+2)^2\,\mbox{,}\\ x^2-10x+25 &= (x-5)^2\,\mbox{.}\end{align*} |
This can be used to solve quadratic equations, for example,
| \displaystyle \begin{align*} x^2+4x+4 &= 9\,\mbox{,}\\ (x+2)^2 &= 9\,\mbox{.}\end{align*} |
Taking roots then gives that \displaystyle x+2=\pm\sqrt{9} and thus that \displaystyle x=-2\pm 3, i.e. \displaystyle x=1 or \displaystyle x=-5.
Sometimes it is necessary to add or subtract an appropriate number to obtain a suitable expression. The above equation, for example, could just as easily been presented to us as
| \displaystyle x^2+4x-5=0\,\mbox{.} |
By adding 9 to both sides, we get a suitable expression on the left side:
| \displaystyle \begin{align*} x^2+4x-5+9 &= 0+9\,\mbox{,}\\ x^2+4x+4\phantom{{}+9} &= 9\,\mbox{.}\end{align*} |
This method is called completing the square.
Example 10
- Solve the equation \displaystyle \ x^2-6x+7=2\,.
The coefficient in front of \displaystyle x is \displaystyle -6 and it shows that we must have the number \displaystyle (-3)^2=9 as the constant term on the left-hand side to make a complete square. By adding \displaystyle 2 to both sides we achieve this:\displaystyle \begin{align*} x^2-6x+7+2 &= 2+2\,\mbox{,}\\ x^2-6x+9\phantom{{}+2} &= 4\,\mbox{,}\\ \rlap{(x-3)^2}\phantom{x^2-6x+7+2}{} &= 4\,\mbox{.}\end{align*} Taking roots then gives \displaystyle x-3=\pm 2, which means that \displaystyle x=1 or \displaystyle x=5.
- Solve the equation \displaystyle \ z^2+21=4-8z\,.
The equation can be written as \displaystyle z^2+8z+17=0. By subtracting 1 on both sides, we get a complete square on the left-hand side:\displaystyle \begin{align*} z^2+8z+17-1 &= 0-1\,\mbox{,}\\ z^2+8z+16\phantom{{}-1} &= -1\,\mbox{,}\\ \rlap{(z+4)^2}\phantom{z^2+8z+17-1}{} &= -1\,\mbox{,}\end{align*} and thus \displaystyle z+4=\pm\sqrt{-1}. In other words, the solutions are \displaystyle z=-4-i and \displaystyle z=-4+i.
Generally, completing the square may be regarded as arranging that "the square of half the coefficient of the x-term" is the constant term in the quadratic expression. This term can always be added to the two sides without worrying about the other terms and then manipulating the equation. If the coefficients of the expression are complex numbers, one still can go about it in the same way.
Example 11
Solve the equation \displaystyle \ x^2-\frac{8}{3}x+1=2\,.
Half the coefficient of \displaystyle x is \displaystyle -\tfrac{4}{3}. We thus add \displaystyle \bigl(-\tfrac{4}{3}\bigr)^2=\tfrac{16}{9} to both sides
| \displaystyle \begin{align*} x^2-\tfrac{8}{3}x+\tfrac{16}{9}+1 &= 2+\tfrac{16}{9}\,\mbox{,}\\ \rlap{\bigl(x-\tfrac{4}{3}\bigr)^2}\phantom{x^2-\tfrac{8}{3}x+\tfrac{16}{9}}{}+1 &= \tfrac{34}{9}\,\mbox{,}\\ \rlap{\bigl(x-\tfrac{4}{3}\bigr)^2}\phantom{x^2-\tfrac{8}{3}x+\tfrac{16}{9}+1} &= \tfrac{25}{9}\,\mbox{.}\end{align*} |
Now it's easy to get to \displaystyle x-\tfrac{4}{3}=\pm\tfrac{5}{3} and thus to get that \displaystyle x=\tfrac{4}{3}\pm\tfrac{5}{3}, i.e. \displaystyle x=-\tfrac{1}{3} or \displaystyle x=3.
Example 12
Solve the equation \displaystyle \ x^2+px+q=0\,.
Completing the square gives
| \displaystyle \begin{align*} x^2+px+\Bigl(\frac{p}{2}\Bigr)^2+q &= \Bigl(\frac{p}{2}\Bigr)^2\,\mbox{,}\\ \rlap{\Bigl(x+\frac{p}{2}\Bigr)^2}\phantom{x^2+px+\Bigl(\frac{p}{2}\Bigr)^2+q}{} &= \Bigl(\frac{p}{2}\Bigr)^2-q\,\mbox{,}\\ \rlap{x+\frac{p}{2}}\phantom{x^2+px+\Bigl(\frac{p}{2}\Bigr)^2+q}{} &= \pm \sqrt{\Bigl(\frac{p}{2}\Bigr)^2-q}\ \mbox{.}\end{align*} |
This gives the usual formula, pq-formula, for solutions to quadratic equations
| \displaystyle x=-\frac{p}{2}\pm \sqrt{\Bigl(\frac{p}{2}\Bigr)^2-q}\,\mbox{.} |
Example 13
Solve the equation \displaystyle \ z^2-(12+4i)z-4+24i=0\,.
Half the coefficient of \displaystyle z is \displaystyle -(6+2i) so we add the square of this expression to both sides
| \displaystyle z^2-(12+4i)z+(-(6+2i))^2-4+24i=(-(6+2i))^2\,\mbox{.} |
Expanding the square on the right-hand side \displaystyle \ (-(6+2i))^2=36+24i+4i^2=32+24i\ and completing the square on the left-hand side gives
| \displaystyle \begin{align*} (z-(6+2i))^2-4+24i &= 32+24i\,\mbox{,}\\ \rlap{(z-(6+2i))^2}\phantom{(z-(6+2i))^2-4+24i}{} &= 36\,\mbox{.}\end{align*} |
After a taking roots, we have that \displaystyle \ z-(6+2i)=\pm 6\ and therefore the solutions are \displaystyle z=12+2i and \displaystyle z=2i.
If one wants to bring about a square in an expression one can use the same technique. In order not to change the value of the expression one both adds and subtracts the missing constant term, such as in the following,
| \displaystyle \begin{align*} x^2+10x+3 &= x^2+10x+25+3-25\\ &= (x+5)^2-22\,\mbox{.}\end{align*} |
Example 14
Complete the square in the expression \displaystyle \ z^2+(2-4i)z+1-3i\,.
Add and subtract the term \displaystyle \bigl(\frac{1}{2}(2-4i)\bigr)^2=(1-2i)^2=-3-4i\,,
| \displaystyle \begin{align*} z^2+(2-4i)z+1-3i &= z^2+(2-4i)z+(1-2i)^2-(1-2i)^2+1-3i\\ &= \bigl(z+(1-2i)\bigr)^2-(1-2i)^2+1-3i\\ &= \bigl(z+(1-2i)\bigr)^2-(-3-4i)+1-3i\\ &= \bigl(z+(1-2i)\bigr)^2+4+i\,\mbox{.}\end{align*} |
Solving using a formula
To solve quadratic equations sometimes the simplest method is to use the usual formula for quadratic equations. However, this may lead to that one ends up with terms of the type \displaystyle \sqrt{a+ib}. One can then assume
| \displaystyle z=x+iy=\sqrt{a+ib}\,\mbox{.} |
By squaring both sides we get
| \displaystyle \begin{align*} (x+iy)^2 &= a+ib\,\mbox{,}\\ x^2 - y^2 + 2xy\,i &= a+ib\,\mbox{.}\end{align*} |
Matching the real and imaginary parts gives
| \displaystyle \left\{\begin{align*} &x^2 - y^2 = a\,\mbox{,}\\ &2xy=b\,\mbox{.}\end{align*}\right. |
These equations can be solved by substitution, for example, \displaystyle y= b/(2x) can be inserted in the first equation.
Example 15
Calculate \displaystyle \ \sqrt{-3-4i}\,.
Put \displaystyle \ x+iy=\sqrt{-3-4i}\ where \displaystyle x and \displaystyle y are real numbers. Squaring both sides gives
| \displaystyle \begin{align*} (x+iy)^2 &= -3-4i\,\mbox{,}\\ x^2 - y^2 + 2xyi &= -3-4i\,\mbox{,}\end{align*} |
which leads to the system of equations
| \displaystyle \Bigl\{\begin{align*} x^2 - y^2 &= -3\,\mbox{,}\\ 2xy&= -4\,\mbox{.}\end{align*} |
From the second equation, we can solve for \displaystyle \ y=-4/(2x) = -2/x\ and put it into the first equation to get
| \displaystyle x^2-\frac{4}{x^2} = -3 \quad \Leftrightarrow \quad x^4 +3x^2 - 4=0\,\mbox{.} |
This is a quadratic equation in \displaystyle x^2 which can be seen more easily by putting \displaystyle t=x^2,
| \displaystyle t^2 +3t -4=0\,\mbox{.} |
The solutions are \displaystyle t = 1 and \displaystyle t = -4. The latter solution must be rejected, as \displaystyle x and \displaystyle y have been assumed to be real numbers, and thus \displaystyle x^2=-4 cannot be true. We get \displaystyle x=\pm\sqrt{1}, which gives us two possible solutions
- \displaystyle \ x=-1\ which gives \displaystyle \ y=-2/(-1)=2\,,
- \displaystyle \ x=1\ which gives \displaystyle \ y=-2/1=-2\,.
So we can conclude that
| \displaystyle \sqrt{-3-4i} = \biggl\{\begin{align*} &\phantom{-}1-2i\,\mbox{,}\\ &-1+2i\,\mbox{.}\end{align*} |
Example 16
- Solve the equation \displaystyle \ z^2-2z+10=0\,.
The formula for solutions to a quadratic equations (see example 3) gives that\displaystyle z= 1\pm \sqrt{1-10} = 1\pm \sqrt{-9}= 1\pm 3i\,\mbox{.} - Solve the equation \displaystyle \ z^2 + (4-2i)z -4i=0\,\mbox{.}
Here, once again , the pq-formula may be used giving the solutions directly\displaystyle \begin{align*} z &= -2+i\pm\sqrt{\smash{(-2+i)^2+4i}\vphantom{i^2}} = -2+i\pm\sqrt{4-4i+i^{\,2}+4i}\\ &=-2+i\pm\sqrt{3} = -2\pm\sqrt{3}+i\,\mbox{.}\end{align*} - Solve the equation \displaystyle \ iz^2+(2+6i)z+2+11i=0\,\mbox{.}
Division of both sides with \displaystyle i gives\displaystyle \begin{align*} z^2 + \frac{2+6i}{i}z +\frac{2+11i}{i} &= 0\,\mbox{,}\\ z^2+ (6-2i)z + 11-2i &= 0\,\mbox{.}\end{align*} Applying the pq- formula gives
\displaystyle \begin{align*} z &= -3+i \pm \sqrt{\smash{(-3+i)^2 -(11-2i)}\vphantom{i^2}}\\ &= -3+i \pm \sqrt{-3-4i}\\ &= -3+i\pm(1-2i)\end{align*} where we used the resulting value of\displaystyle \ \sqrt{-3-4i}\ which we obtained in example 15. The solutions are therefore
\displaystyle z=\biggl\{\begin{align*} &-2-i\,\mbox{,}\\ &-4+3i\,\mbox{.}\end{align*}
