Whakaoti i te 1/2-i

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

Aromātai

$$\frac{2}{5}+\frac{1}{5}i=0.4+0.2i$$

Ngā Upane Otinga

Whakareatia te taurunga me te tauraro ki te haumi hiato o te tauraro, 2+i.

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

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}.

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

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

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

Whakareatia te 1 ki te 2+i, ka 2+i.

$$\frac{2+i}{5}$$

Whakawehea te 2+i ki te 5, kia riro ko \frac{2}{5}+\frac{1}{5}i.

$$\frac{2}{5}+\frac{1}{5}i$$

Wāhi Tūturu

$$\frac{2}{5} = 0.4$$

Ngā Upane Otinga

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

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

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}.

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

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

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

Whakareatia te 1 ki te 2+i, ka 2+i.

$$Re(\frac{2+i}{5})$$

Whakawehea te 2+i ki te 5, kia riro ko \frac{2}{5}+\frac{1}{5}i.

$$Re(\frac{2}{5}+\frac{1}{5}i)$$

Ko te wāhi tūturu o \frac{2}{5}+\frac{1}{5}i ko \frac{2}{5}.

$$\frac{2}{5}$$