Whakaoti i te 3i(1-i)/1+i

$$\frac { 3 i ( 1 - i ) } { 1 + i }$$

Aromātai

$$3$$

Ngā Upane Otinga

Whakareatia 3i ki te 1-i.

$$\frac{3i\times 1+3\left(-1\right)i^{2}}{1+i}$$

Hei tōna tikanga, ko te i^{2} ko -1.

$$\frac{3i\times 1+3\left(-1\right)\left(-1\right)}{1+i}$$

Mahia ngā whakarea i roto o 3i\times 1+3\left(-1\right)\left(-1\right). Whakaraupapatia anō ngā kīanga tau.

$$\frac{3+3i}{1+i}$$

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

$$\frac{\left(3+3i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}$$

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}.

$$\frac{\left(3+3i\right)\left(1-i\right)}{1^{2}-i^{2}}$$

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

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

Me whakarea ngā tau matatini 3+3i me 1-i pēnā i te whakarea huarua.

$$\frac{3\times 1+3\left(-i\right)+3i\times 1+3\left(-1\right)i^{2}}{2}$$

Hei tōna tikanga, ko te i^{2} ko -1.

$$\frac{3\times 1+3\left(-i\right)+3i\times 1+3\left(-1\right)\left(-1\right)}{2}$$

Mahia ngā whakarea i roto o 3\times 1+3\left(-i\right)+3i\times 1+3\left(-1\right)\left(-1\right).

$$\frac{3-3i+3i+3}{2}$$

Whakakotahitia ngā wāhi tūturu me ngā wāhi pōhewa ki 3-3i+3i+3.

$$\frac{3+3+\left(-3+3\right)i}{2}$$

Mahia ngā tāpiri i roto o 3+3+\left(-3+3\right)i.

$$\frac{6}{2}$$

Whakawehea te 6 ki te 2, kia riro ko 3.

$$3$$

Wāhi Tūturu

$$3$$

Ngā Upane Otinga

Whakareatia 3i ki te 1-i.

$$Re(\frac{3i\times 1+3\left(-1\right)i^{2}}{1+i})$$

Hei tōna tikanga, ko te i^{2} ko -1.

$$Re(\frac{3i\times 1+3\left(-1\right)\left(-1\right)}{1+i})$$

Mahia ngā whakarea i roto o 3i\times 1+3\left(-1\right)\left(-1\right). Whakaraupapatia anō ngā kīanga tau.

$$Re(\frac{3+3i}{1+i})$$

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

$$Re(\frac{\left(3+3i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)})$$

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}.

$$Re(\frac{\left(3+3i\right)\left(1-i\right)}{1^{2}-i^{2}})$$

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

$$Re(\frac{\left(3+3i\right)\left(1-i\right)}{2})$$

Me whakarea ngā tau matatini 3+3i me 1-i pēnā i te whakarea huarua.

$$Re(\frac{3\times 1+3\left(-i\right)+3i\times 1+3\left(-1\right)i^{2}}{2})$$

Hei tōna tikanga, ko te i^{2} ko -1.

$$Re(\frac{3\times 1+3\left(-i\right)+3i\times 1+3\left(-1\right)\left(-1\right)}{2})$$

Mahia ngā whakarea i roto o 3\times 1+3\left(-i\right)+3i\times 1+3\left(-1\right)\left(-1\right).

$$Re(\frac{3-3i+3i+3}{2})$$

Whakakotahitia ngā wāhi tūturu me ngā wāhi pōhewa ki 3-3i+3i+3.

$$Re(\frac{3+3+\left(-3+3\right)i}{2})$$

Mahia ngā tāpiri i roto o 3+3+\left(-3+3\right)i.

$$Re(\frac{6}{2})$$

Whakawehea te 6 ki te 2, kia riro ko 3.

$$Re(3)$$

Ko te wāhi tūturu o 3 ko 3.

$$3$$