Whakaoti i te 2i/1+i

$$\frac { 2 i } { 1 + i }$$

Aromātai

$$1+i$$

Ngā Upane Otinga

Whakareatia te taurunga me te tauraro ki te haumi hiato o te tauraro, 1-i.

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

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}.

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

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

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

Whakareatia 2i ki te 1-i.

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

Hei tōna tikanga, ko te i^{2} ko -1.

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

Mahia ngā whakarea i roto o 2i\times 1+2\left(-1\right)\left(-1\right). Whakaraupapatia anō ngā kīanga tau.

$$\frac{2+2i}{2}$$

Whakawehea te 2+2i ki te 2, kia riro ko 1+i.

$$1+i$$

Wāhi Tūturu

$$1$$

Ngā Upane Otinga

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

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

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}.

$$Re(\frac{2i\left(1-i\right)}{1^{2}-i^{2}})$$

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

$$Re(\frac{2i\left(1-i\right)}{2})$$

Whakareatia 2i ki te 1-i.

$$Re(\frac{2i\times 1+2\left(-1\right)i^{2}}{2})$$

Hei tōna tikanga, ko te i^{2} ko -1.

$$Re(\frac{2i\times 1+2\left(-1\right)\left(-1\right)}{2})$$

Mahia ngā whakarea i roto o 2i\times 1+2\left(-1\right)\left(-1\right). Whakaraupapatia anō ngā kīanga tau.

$$Re(\frac{2+2i}{2})$$

Whakawehea te 2+2i ki te 2, kia riro ko 1+i.

$$Re(1+i)$$

Ko te wāhi tūturu o 1+i ko 1.

$$1$$