Whakaoti i te 3+xi/1+2i=y+2i

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

Whakaoti mō ξ

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

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakawehea ia wā o 3+\xi ki te 1+2i, kia riro ko \frac{3}{1+2i}+\frac{\xi }{1+2i}.

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

Me whakarea te taurunga me te tauraro o \frac{3}{1+2i} ki te haumi hiato o te tauraro, 1-2i.

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

Mahia ngā whakarea i roto o \frac{3\left(1-2i\right)}{\left(1+2i\right)\left(1-2i\right)}.

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

Whakawehea te 3-6i ki te 5, kia riro ko \frac{3}{5}-\frac{6}{5}i.

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

Tangohia te \frac{3}{5}-\frac{6}{5}i mai i ngā taha e rua.

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

Whakareatia te -1 ki te \frac{3}{5}-\frac{6}{5}i, ka -\frac{3}{5}+\frac{6}{5}i.

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

Mahia ngā tāpiri i roto o 2i+\left(-\frac{3}{5}+\frac{6}{5}i\right).

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

He hanga arowhānui tō te whārite.

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

Whakawehea ngā taha e rua ki te \frac{1}{5}-\frac{2}{5}i.

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

Mā te whakawehe ki te \frac{1}{5}-\frac{2}{5}i ka wetekia te whakareanga ki te \frac{1}{5}-\frac{2}{5}i.

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

Whakawehe y+\left(-\frac{3}{5}+\frac{16}{5}i\right) ki te \frac{1}{5}-\frac{2}{5}i.

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

Whakaoti mō y

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