Aus Online Mathematik Brückenkurs 2

Version vom 11:38, 31. Okt. 2008 (bearbeiten) Tek (Diskussion | Beiträge) K ← Zum vorherigen Versionsunterschied		Version vom 13:14, 10. Mär. 2009 (bearbeiten) (rückgängig) Tekbot (Diskussion | Beiträge) K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
	Let us first divide both sides by <math>4+i</math>, so that the coefficient in front of <math>z^2</math> becomes <math>1</math>,		Let us first divide both sides by <math>4+i</math>, so that the coefficient in front of <math>z^2</math> becomes <math>1</math>,
-	{{Displayed math||<math>z^2 + \frac{1-21i}{4+i}z = \frac{17}{4+i}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<math>z^2 + \frac{1-21i}{4+i}z = \frac{17}{4+i}\,\textrm{.}</math>}}
	The two complex quotients become		The two complex quotients become
-	{{Displayed math||<math>\begin{align}	+	{{Abgesetzte Formel||<math>\begin{align}
	\frac{1-21i}{4+i}		\frac{1-21i}{4+i}
	&= \frac{(1-21i)(4-i)}{(4+i)(4-i)}		&= \frac{(1-21i)(4-i)}{(4+i)(4-i)}
Zeile 21:		Zeile 21:
	Thus, the equation becomes		Thus, the equation becomes
-	{{Displayed math||<math>z^2 - (1+5i)z = 4-i\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<math>z^2 - (1+5i)z = 4-i\,\textrm{.}</math>}}
	Now, we complete the square of the left-hand side,		Now, we complete the square of the left-hand side,
-	{{Displayed math||<math>\begin{align}	+	{{Abgesetzte Formel||<math>\begin{align}
	\Bigl(z-\frac{1+5i}{2}\Bigr)^2 - \Bigl(\frac{1+5i}{2}\Bigr)^2 &= 4-i\,,\\[5pt]		\Bigl(z-\frac{1+5i}{2}\Bigr)^2 - \Bigl(\frac{1+5i}{2}\Bigr)^2 &= 4-i\,,\\[5pt]
	\Bigl(z-\frac{1+5i}{2}\Bigr)^2 - \Bigl(\frac{1}{4}+\frac{5}{2}\,i+\frac{25}{4}i^2 \Bigr) &= 4-i\,,\\[5pt]		\Bigl(z-\frac{1+5i}{2}\Bigr)^2 - \Bigl(\frac{1}{4}+\frac{5}{2}\,i+\frac{25}{4}i^2 \Bigr) &= 4-i\,,\\[5pt]
Zeile 34:		Zeile 34:
	If we set <math>w=z-\frac{1+5i}{2}</math>, we have a binomial equation in <math>w</math>,		If we set <math>w=z-\frac{1+5i}{2}</math>, we have a binomial equation in <math>w</math>,
-	{{Displayed math||<math>w^2 = -2+\frac{3}{2}\,i</math>}}	+	{{Abgesetzte Formel||<math>w^2 = -2+\frac{3}{2}\,i</math>}}
	which we solve by putting <math>w=x+iy</math>,		which we solve by putting <math>w=x+iy</math>,
-	{{Displayed math||<math>(x+iy)^2 = -2+\frac{3}{2}\,i</math>}}	+	{{Abgesetzte Formel||<math>(x+iy)^2 = -2+\frac{3}{2}\,i</math>}}
	or, if the left-hand side is expanded,		or, if the left-hand side is expanded,
-	{{Displayed math||<math>x^2-y^2+2xyi = -2+\frac{3}{2}\,i\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<math>x^2-y^2+2xyi = -2+\frac{3}{2}\,i\,\textrm{.}</math>}}
	If we identify the real and imaginary parts on both sides, we get		If we identify the real and imaginary parts on both sides, we get
-	{{Displayed math||<math>\begin{align}	+	{{Abgesetzte Formel||<math>\begin{align}
	x^2-y^2 &= -2\,,\\[5pt]		x^2-y^2 &= -2\,,\\[5pt]
	2xy &= \frac{3}{2}\,,		2xy &= \frac{3}{2}\,,
Zeile 53:		Zeile 53:
	and if we take the magnitude of both sides and put them equal to each other, we obtain a third relation:		and if we take the magnitude of both sides and put them equal to each other, we obtain a third relation:
-	{{Displayed math||<math>x^2 + y^2 = \sqrt{(-2)^2+\bigl(\tfrac{3}{2}\bigr)^2} = \tfrac{5}{2}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<math>x^2 + y^2 = \sqrt{(-2)^2+\bigl(\tfrac{3}{2}\bigr)^2} = \tfrac{5}{2}\,\textrm{.}</math>}}
	Strictly speaking, this last relation is contained within the first two and we include it because makes the calculations easier.		Strictly speaking, this last relation is contained within the first two and we include it because makes the calculations easier.
Zeile 59:		Zeile 59:
	Together, the three relations constitute the following system of equations:		Together, the three relations constitute the following system of equations:
-	{{Displayed math||<math>\left\{\begin{align}	+	{{Abgesetzte Formel||<math>\left\{\begin{align}
	x^2-y^2 &= -2\,,\\[5pt]		x^2-y^2 &= -2\,,\\[5pt]
	2xy &= \frac{3}{2}\,,\\[5pt]		2xy &= \frac{3}{2}\,,\\[5pt]
Zeile 125:		Zeile 125:
	This gives four possible combinations,		This gives four possible combinations,
-	{{Displayed math||<math>\left\{\begin{align}	+	{{Abgesetzte Formel||<math>\left\{\begin{align}
	x &= \tfrac{1}{2}\\[5pt]		x &= \tfrac{1}{2}\\[5pt]
	y &= \tfrac{3}{2}		y &= \tfrac{3}{2}
Zeile 147:		Zeile 147:
	of which only two also satisfy the second equation.		of which only two also satisfy the second equation.
-	{{Displayed math||<math>\left\{\begin{align}	+	{{Abgesetzte Formel||<math>\left\{\begin{align}
	x &= \tfrac{1}{2}\\[5pt]		x &= \tfrac{1}{2}\\[5pt]
	y &= \tfrac{3}{2}		y &= \tfrac{3}{2}
Zeile 159:		Zeile 159:
	This means that the binomial equation has the two solutions,		This means that the binomial equation has the two solutions,
-	{{Displayed math||<math>w=\frac{1}{2}+\frac{3}{2}\,i\qquad</math> and <math>\qquad w=-\frac{1}{2}-\frac{3}{2}\,i\,,</math>}}	+	{{Abgesetzte Formel||<math>w=\frac{1}{2}+\frac{3}{2}\,i\qquad</math> and <math>\qquad w=-\frac{1}{2}-\frac{3}{2}\,i\,,</math>}}
	and that the original equation has the solutions		and that the original equation has the solutions
-	{{Displayed math||<math>z=1+4i\qquad</math> and <math>\qquad z=i</math>}}	+	{{Abgesetzte Formel||<math>z=1+4i\qquad</math> and <math>\qquad z=i</math>}}
	according to the relation <math>w=z-\frac{1+5i}{2}</math>.		according to the relation <math>w=z-\frac{1+5i}{2}</math>.

---

Version vom 13:14, 10. Mär. 2009

Let us first divide both sides by \displaystyle 4+i, so that the coefficient in front of \displaystyle z^2 becomes \displaystyle 1,

\displaystyle z^2 + \frac{1-21i}{4+i}z = \frac{17}{4+i}\,\textrm{.}

The two complex quotients become

\displaystyle \begin{align} \frac{1-21i}{4+i} &= \frac{(1-21i)(4-i)}{(4+i)(4-i)} = \frac{4-i-84i+21i^2}{4^2-i^2}\\[5pt] &= \frac{-17-85i}{16+1} = \frac{-17-85i}{17} = -1-5i\,,\\[10pt] \frac{17}{4+i} &= \frac{17(4-i)}{(4+i)(4-i)} = \frac{17(4-i)}{4^2-i^2}\\[5pt] &= \frac{17(4-i)}{17} = 4-i\,\textrm{.} \end{align}

Thus, the equation becomes

\displaystyle z^2 - (1+5i)z = 4-i\,\textrm{.}

Now, we complete the square of the left-hand side,

\displaystyle \begin{align} \Bigl(z-\frac{1+5i}{2}\Bigr)^2 - \Bigl(\frac{1+5i}{2}\Bigr)^2 &= 4-i\,,\\[5pt] \Bigl(z-\frac{1+5i}{2}\Bigr)^2 - \Bigl(\frac{1}{4}+\frac{5}{2}\,i+\frac{25}{4}i^2 \Bigr) &= 4-i\,,\\[5pt] \Bigl(z-\frac{1+5i}{2}\Bigr)^2 - \frac{1}{4} - \frac{5}{2}i + \frac{25}{4} &= 4-i\,, \\[5pt] \Bigl(z-\frac{1+5i}{2}\Bigr)^2 &= -2+\frac{3}{2}\,i\,\textrm{.} \end{align}

If we set \displaystyle w=z-\frac{1+5i}{2}, we have a binomial equation in \displaystyle w,

\displaystyle w^2 = -2+\frac{3}{2}\,i

which we solve by putting \displaystyle w=x+iy,

\displaystyle (x+iy)^2 = -2+\frac{3}{2}\,i

or, if the left-hand side is expanded,

\displaystyle x^2-y^2+2xyi = -2+\frac{3}{2}\,i\,\textrm{.}

If we identify the real and imaginary parts on both sides, we get

\displaystyle \begin{align} x^2-y^2 &= -2\,,\\[5pt] 2xy &= \frac{3}{2}\,, \end{align}

and if we take the magnitude of both sides and put them equal to each other, we obtain a third relation:

\displaystyle x^2 + y^2 = \sqrt{(-2)^2+\bigl(\tfrac{3}{2}\bigr)^2} = \tfrac{5}{2}\,\textrm{.}

Strictly speaking, this last relation is contained within the first two and we include it because makes the calculations easier.

Together, the three relations constitute the following system of equations:

\displaystyle \left\{\begin{align} x^2-y^2 &= -2\,,\\[5pt] 2xy &= \frac{3}{2}\,,\\[5pt] x^2 + y^2 &= \frac{5}{2}\,\textrm{.} \end{align} \right.

From the first and the third equations, we can relatively easily obtain the values that \displaystyle x and \displaystyle y can take.

Add the first and third equations,

	\displaystyle x^2	\displaystyle {}-{}	\displaystyle y^2	\displaystyle {}={}	\displaystyle -2
\displaystyle +\ \	\displaystyle x^2	\displaystyle {}+{}	\displaystyle y^2	\displaystyle {}={}	\displaystyle \tfrac{5}{2}
	\displaystyle 2x^2			\displaystyle {}={}	\displaystyle \tfrac{1}{2}

which gives that \displaystyle x=\pm \tfrac{1}{2}.

Then, subtract the first equation from the third equation,

	\displaystyle x^2	\displaystyle {}+{}	\displaystyle y^2	\displaystyle {}={}	\displaystyle \tfrac{5}{2}
\displaystyle -\ \	\displaystyle \bigl(x^2	\displaystyle {}-{}	\displaystyle y^2	\displaystyle {}={}	\displaystyle -2\rlap{\bigr)}
			\displaystyle 2y^2	\displaystyle {}={}	\displaystyle \tfrac{9}{2}

i.e. \displaystyle y=\pm\tfrac{3}{2}.

This gives four possible combinations,

\displaystyle \left\{\begin{align} x &= \tfrac{1}{2}\\[5pt] y &= \tfrac{3}{2} \end{align}\right. \qquad \left\{\begin{align} x &= \tfrac{1}{2}\\[5pt] y &= -\tfrac{3}{2} \end{align}\right. \qquad \left\{\begin{align} x &= -\tfrac{1}{2}\\[5pt] y &= \tfrac{3}{2} \end{align}\right. \qquad \left\{\begin{align} x &= -\tfrac{1}{2}\\[5pt] y &= -\tfrac{3}{2} \end{align} \right.

of which only two also satisfy the second equation.

\displaystyle \left\{\begin{align} x &= \tfrac{1}{2}\\[5pt] y &= \tfrac{3}{2} \end{align}\right. \qquad\text{and}\qquad \left\{\begin{align} x &= -\tfrac{1}{2}\\[5pt] y &= -\tfrac{3}{2} \end{align}\right.

This means that the binomial equation has the two solutions,

\displaystyle w=\frac{1}{2}+\frac{3}{2}\,i\qquad and \displaystyle \qquad w=-\frac{1}{2}-\frac{3}{2}\,i\,,

and that the original equation has the solutions

\displaystyle z=1+4i\qquad and \displaystyle \qquad z=i

according to the relation \displaystyle w=z-\frac{1+5i}{2}.

Finally, we check that the solutions really do satisfy the equation.

\displaystyle \begin{align} z=1+4i:\quad (4+i)z^2+(1-21i)z &= (4+i)(1+4i)^2+(1-21i)(1+4i)\\[5pt] &= (4+i)(1+8i+16i^2)+(1+4i-21i-84i^2)\\[5pt] &= (4+i)(-15+8i)+1-17i+84\\[5pt] &= -60+32i-15i+8i^2+1-17i+84\\[5pt] &= -60+32i-15i-8+1-17i+84\\[5pt] &= 17\,,\\[10pt] z={}\rlap{i:}\phantom{1+4i:}{}\quad (4+i)z^2+(1-21i)z &= (4+i)i^2 + (1-21i)i\\[5pt] &= (4+i)(-1)+i-21i^2\\[5pt] &= -4-i+i+21\\[5pt] &= 17\,\textrm{.} \end{align}