Løs 3+xi/1+2i=y+2i

$$\frac { 3 + x i } { 1 + 2 i } = y + 2 i$$

Løs for ξ

$$\xi =\left(1+2i\right)y+\left(-7+2i\right)$$

Vejledning i at løse lineær ligning

Divider hvert led på 3+\xi med 1+2i for at få \frac{3}{1+2i}+\frac{\xi }{1+2i}.

$$\frac{3}{1+2i}+\frac{\xi }{1+2i}=y+2i$$

Multiplicer både tælleren og nævneren af \frac{3}{1+2i} med nævnerens komplekse konjugation, 1-2i.

$$\frac{3\left(1-2i\right)}{\left(1+2i\right)\left(1-2i\right)}+\frac{\xi }{1+2i}=y+2i$$

Lav multiplikationerne i \frac{3\left(1-2i\right)}{\left(1+2i\right)\left(1-2i\right)}.

$$\frac{3-6i}{5}+\frac{\xi }{1+2i}=y+2i$$

Divider 3-6i med 5 for at få \frac{3}{5}-\frac{6}{5}i.

$$\frac{3}{5}-\frac{6}{5}i+\frac{\xi }{1+2i}=y+2i$$

Subtraher \frac{3}{5}-\frac{6}{5}i fra begge sider.

$$\frac{\xi }{1+2i}=y+2i-\left(\frac{3}{5}-\frac{6}{5}i\right)$$

Multiplicer -1 og \frac{3}{5}-\frac{6}{5}i for at få -\frac{3}{5}+\frac{6}{5}i.

$$\frac{\xi }{1+2i}=y+2i+\left(-\frac{3}{5}+\frac{6}{5}i\right)$$

Lav additionerne i 2i+\left(-\frac{3}{5}+\frac{6}{5}i\right).

$$\frac{\xi }{1+2i}=y-\frac{3}{5}+\frac{16}{5}i$$

Ligningen er nu i standardform.

$$\left(\frac{1}{5}-\frac{2}{5}i\right)\xi =y+\left(-\frac{3}{5}+\frac{16}{5}i\right)$$

Divider begge sider med \frac{1}{5}-\frac{2}{5}i.

$$\frac{\left(\frac{1}{5}-\frac{2}{5}i\right)\xi }{\frac{1}{5}-\frac{2}{5}i}=\frac{y+\left(-\frac{3}{5}+\frac{16}{5}i\right)}{\frac{1}{5}-\frac{2}{5}i}$$

Division med \frac{1}{5}-\frac{2}{5}i annullerer multiplikationen med \frac{1}{5}-\frac{2}{5}i.

$$\xi =\frac{y+\left(-\frac{3}{5}+\frac{16}{5}i\right)}{\frac{1}{5}-\frac{2}{5}i}$$

Divider y+\left(-\frac{3}{5}+\frac{16}{5}i\right) med \frac{1}{5}-\frac{2}{5}i.

$$\xi =\left(1+2i\right)y+\left(-7+2i\right)$$

Løs for y

$$y=\left(\frac{1}{5}-\frac{2}{5}i\right)\xi +\left(\frac{3}{5}-\frac{16}{5}i\right)$$