Whakaoti i te (2-i)z=1+i.ineqj

$$( 2 - i ) z = 1 + i . i \neq j$$

Whakaoti mō z

$$z=\frac{1}{5}+\frac{3}{5}i=0.2+0.6i$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakawehea ngā taha e rua ki te 2-i.

$$z=\frac{1+i}{2-i}$$

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

$$z=\frac{\left(1+i\right)\left(2+i\right)}{\left(2-i\right)\left(2+i\right)}$$

Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$z=\frac{\left(1+i\right)\left(2+i\right)}{2^{2}-i^{2}}$$

Hei tōna tikanga, ko te i^{2} ko -1. Tātaitia te tauraro.

$$z=\frac{\left(1+i\right)\left(2+i\right)}{5}$$

Me whakarea ngā tau matatini 1+i me 2+i pēnā i te whakarea huarua.

$$z=\frac{1\times 2+i+2i+i^{2}}{5}$$

Hei tōna tikanga, ko te i^{2} ko -1.

$$z=\frac{1\times 2+i+2i-1}{5}$$

Mahia ngā whakarea i roto o 1\times 2+i+2i-1.

$$z=\frac{2+i+2i-1}{5}$$

Whakakotahitia ngā wāhi tūturu me ngā wāhi pōhewa ki 2+i+2i-1.

$$z=\frac{2-1+\left(1+2\right)i}{5}$$

Mahia ngā tāpiri i roto o 2-1+\left(1+2\right)i.

$$z=\frac{1+3i}{5}$$

Whakawehea te 1+3i ki te 5, kia riro ko \frac{1}{5}+\frac{3}{5}i.

$$z=\frac{1}{5}+\frac{3}{5}i$$