From Förberedande kurs i matematik 2

Revision as of 10:54, 7 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:3_1_4e-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:3_1_4e-2(2).gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (08:20, 30 October 2008) (edit) (undo) Tek (Talk | contribs) m
(3 intermediate revisions not shown.)
Line 1:		Line 1:
-	{{NAVCONTENT_START}}	+	If we multiply both sides by <math>z+i</math>, we avoid having <math>z</math> in the denominator,
-	<center> [[Bild:3_1_4e-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>iz+1=(3+i)(z+i)\,\textrm{.}</math>}}
-	{{NAVCONTENT_START}}	+
-	<center> [[Bild:3_1_4e-2(2).gif]] </center>	+	At the same time, this means that we are now working with a new equation which is not necessarily entirely equivalent with the original equation. Should the new equation show itself to have <math>z=-i</math> as a solution, then we must ignore that solution, because our initial equation cannot possibly have <math>z=-i</math> as a solution (the denominator of the left-hand side becomes zero).
-	{{NAVCONTENT_STOP}}	+
		+	We expand the right-hand side in the new equation,
		+
		+	{{Displayed math||<math>iz+1 = 3z+3i+iz-1\,,</math>}}
		+
		+	and move all the terms in <math>z</math> over to the left-hand side and the constants to the right-hand side,
		+
		+	{{Displayed math||<math>\begin{align}
		+	iz-3z-iz &= 3i-1-1\,,\\[5pt]
		+	-3z &= -2+3i\,\textrm{.}
		+	\end{align}</math>}}
		+
		+	Then, we obtain
		+
		+	{{Displayed math||<math>z = \frac{-2+3i}{-3} = \frac{2}{3}-i\,\textrm{.}</math>}}
		+
		+	It is a little troublesome to divide two complex numbers, so we will therefore not check whether <math>z=\tfrac{2}{3}-i</math> is a solution to the original equation, but satisfy ourselves with substituting into the equation <math>iz+1=(3+i)(z+i)</math>,
		+
		+	{{Displayed math||<math>\begin{align}
		+	\text{LHS} &= iz+1 = i(\tfrac{2}{3}-i)+1 = \tfrac{2}{3}\cdot i+1+1 = 2+\tfrac{2}{3}i,\\[5pt]
		+	\text{RHS} &= (3+i)(z+i) = (3+i)(\tfrac{2}{3}-i+i) = (3+i)\tfrac{2}{3} = 2+\tfrac{2}{3}i\,\textrm{.}
		+	\end{align}</math>}}

---

Current revision

If we multiply both sides by \displaystyle z+i, we avoid having \displaystyle z in the denominator,

\displaystyle iz+1=(3+i)(z+i)\,\textrm{.}

At the same time, this means that we are now working with a new equation which is not necessarily entirely equivalent with the original equation. Should the new equation show itself to have \displaystyle z=-i as a solution, then we must ignore that solution, because our initial equation cannot possibly have \displaystyle z=-i as a solution (the denominator of the left-hand side becomes zero).

We expand the right-hand side in the new equation,

\displaystyle iz+1 = 3z+3i+iz-1\,,

and move all the terms in \displaystyle z over to the left-hand side and the constants to the right-hand side,

\displaystyle \begin{align} iz-3z-iz &= 3i-1-1\,,\\[5pt] -3z &= -2+3i\,\textrm{.} \end{align}

Then, we obtain

\displaystyle z = \frac{-2+3i}{-3} = \frac{2}{3}-i\,\textrm{.}

It is a little troublesome to divide two complex numbers, so we will therefore not check whether \displaystyle z=\tfrac{2}{3}-i is a solution to the original equation, but satisfy ourselves with substituting into the equation \displaystyle iz+1=(3+i)(z+i),

\displaystyle \begin{align} \text{LHS} &= iz+1 = i(\tfrac{2}{3}-i)+1 = \tfrac{2}{3}\cdot i+1+1 = 2+\tfrac{2}{3}i,\\[5pt] \text{RHS} &= (3+i)(z+i) = (3+i)(\tfrac{2}{3}-i+i) = (3+i)\tfrac{2}{3} = 2+\tfrac{2}{3}i\,\textrm{.} \end{align}