Whakaoti i te 1+i/z=1-i

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

Whakaoti mō z

$$z=i$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tē taea kia ōrite te tāupe z ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te z.

$$1+i=\left(1-i\right)z$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$\left(1-i\right)z=1+i$$

Whakawehea ngā taha e rua ki te 1-i.

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

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

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

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}.

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

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

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

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

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

Hei tōna tikanga, ko te i^{2} ko -1.

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

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

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

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

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

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

$$z=\frac{2i}{2}$$

Whakawehea te 2i ki te 2, kia riro ko i.

$$z=i$$