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Aus Online Mathematik Brückenkurs 2

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Version vom 10:49, 11. Mär. 2009

Let us first divide both sides by 4+i, so that the coefficient in front of z2 becomes 1,

z2+4+i1−21iz=174+i.

The two complex quotients become

4+i1−21i174+i=(4+i)(4−i)(1−21i)(4−i)=42−i24−i−84i+21i2=16+1−17−85i=17−17−85i=−1−5i=17(4−i)(4+i)(4−i)=42−i217(4−i)=1717(4−i)=4−i.

Thus, the equation becomes

z2−(1+5i)z=4−i.

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

z−21+5i2−21+5i2z−21+5i2−41+25i+425i2z−21+5i2−41−25i+425z−21+5i2=4−i=4−i=4−i=−2+23i.

If we set w=z−21+5i, we have a binomial equation in w,

w2=−2+23i

which we solve by putting w=x+iy,

(x+iy)2=−2+23i

or, if the left-hand side is expanded,

x2−y2+2xyi=−2+23i.

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

x2−y22xy=−2=23

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

x2+y2=(−2)2+232=25.

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:

x2−y22xyx2+y2=−2=23=25.

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

Add the first and third equations,

	x2	−	y2	=	−2
+	x2	+	y2	=	25
	2x2			=	21

which gives that x=21.

Then, subtract the first equation from the third equation,

	x2	+	y2	=	25
−	x2	−	y2	=	−2
			2y2	=	29

i.e. y=23.

This gives four possible combinations,

xy=21=23xy=21=−23xy=−21=23xy=−21=−23

of which only two also satisfy the second equation.

xy=21=23andxy=−21=−23

This means that the binomial equation has the two solutions,

w=21+23i and w=−21−23i

and that the original equation has the solutions

z=1+4i and z=i

according to the relation w=z−21+5i.

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

z=1+4i:(4+i)z2+(1−21i)zz=i:(4+i)z2+(1−21i)z=(4+i)(1+4i)2+(1−21i)(1+4i)=(4+i)(1+8i+16i2)+(1+4i−21i−84i2)=(4+i)(−15+8i)+1−17i+84=−60+32i−15i+8i2+1−17i+84=−60+32i−15i−8+1−17i+84=17=(4+i)i2+(1−21i)i=(4+i)(−1)+i−21i2=−4−i+i+21=17.