Whakaoti i te (3+4i)(1+2i)/1+i=

$$\frac { ( 3 + 4 i ) ( 1 + 2 i ) } { 1 + i } =$$

Aromātai

$$\frac{5}{2}+\frac{15}{2}i=2.5+7.5i$$

Ngā Upane Otinga

Me whakarea ngā tau matatini 3+4i me 1+2i pēnā i te whakarea huarua.

$$\frac{3\times 1+3\times \left(2i\right)+4i\times 1+4\times 2i^{2}}{1+i}$$

Hei tōna tikanga, ko te i^{2} ko -1.

$$\frac{3\times 1+3\times \left(2i\right)+4i\times 1+4\times 2\left(-1\right)}{1+i}$$

Mahia ngā whakarea i roto o 3\times 1+3\times \left(2i\right)+4i\times 1+4\times 2\left(-1\right).

$$\frac{3+6i+4i-8}{1+i}$$

Whakakotahitia ngā wāhi tūturu me ngā wāhi pōhewa ki 3+6i+4i-8.

$$\frac{3-8+\left(6+4\right)i}{1+i}$$

Mahia ngā tāpiri i roto o 3-8+\left(6+4\right)i.

$$\frac{-5+10i}{1+i}$$

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

$$\frac{\left(-5+10i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}$$

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}.

$$\frac{\left(-5+10i\right)\left(1-i\right)}{1^{2}-i^{2}}$$

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

$$\frac{\left(-5+10i\right)\left(1-i\right)}{2}$$

Me whakarea ngā tau matatini -5+10i me 1-i pēnā i te whakarea huarua.

$$\frac{-5-5\left(-i\right)+10i\times 1+10\left(-1\right)i^{2}}{2}$$

Hei tōna tikanga, ko te i^{2} ko -1.

$$\frac{-5-5\left(-i\right)+10i\times 1+10\left(-1\right)\left(-1\right)}{2}$$

Mahia ngā whakarea i roto o -5-5\left(-i\right)+10i\times 1+10\left(-1\right)\left(-1\right).

$$\frac{-5+5i+10i+10}{2}$$

Whakakotahitia ngā wāhi tūturu me ngā wāhi pōhewa ki -5+5i+10i+10.

$$\frac{-5+10+\left(5+10\right)i}{2}$$

Mahia ngā tāpiri i roto o -5+10+\left(5+10\right)i.

$$\frac{5+15i}{2}$$

Whakawehea te 5+15i ki te 2, kia riro ko \frac{5}{2}+\frac{15}{2}i.

$$\frac{5}{2}+\frac{15}{2}i$$

Wāhi Tūturu

$$\frac{5}{2} = 2\frac{1}{2} = 2.5$$

Ngā Upane Otinga

Me whakarea ngā tau matatini 3+4i me 1+2i pēnā i te whakarea huarua.

$$Re(\frac{3\times 1+3\times \left(2i\right)+4i\times 1+4\times 2i^{2}}{1+i})$$

Hei tōna tikanga, ko te i^{2} ko -1.

$$Re(\frac{3\times 1+3\times \left(2i\right)+4i\times 1+4\times 2\left(-1\right)}{1+i})$$

Mahia ngā whakarea i roto o 3\times 1+3\times \left(2i\right)+4i\times 1+4\times 2\left(-1\right).

$$Re(\frac{3+6i+4i-8}{1+i})$$

Whakakotahitia ngā wāhi tūturu me ngā wāhi pōhewa ki 3+6i+4i-8.

$$Re(\frac{3-8+\left(6+4\right)i}{1+i})$$

Mahia ngā tāpiri i roto o 3-8+\left(6+4\right)i.

$$Re(\frac{-5+10i}{1+i})$$

Me whakarea te taurunga me te tauraro o \frac{-5+10i}{1+i} ki te haumi hiato o te tauraro, 1-i.

$$Re(\frac{\left(-5+10i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)})$$

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}.

$$Re(\frac{\left(-5+10i\right)\left(1-i\right)}{1^{2}-i^{2}})$$

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

$$Re(\frac{\left(-5+10i\right)\left(1-i\right)}{2})$$

Me whakarea ngā tau matatini -5+10i me 1-i pēnā i te whakarea huarua.

$$Re(\frac{-5-5\left(-i\right)+10i\times 1+10\left(-1\right)i^{2}}{2})$$

Hei tōna tikanga, ko te i^{2} ko -1.

$$Re(\frac{-5-5\left(-i\right)+10i\times 1+10\left(-1\right)\left(-1\right)}{2})$$

Mahia ngā whakarea i roto o -5-5\left(-i\right)+10i\times 1+10\left(-1\right)\left(-1\right).

$$Re(\frac{-5+5i+10i+10}{2})$$

Whakakotahitia ngā wāhi tūturu me ngā wāhi pōhewa ki -5+5i+10i+10.

$$Re(\frac{-5+10+\left(5+10\right)i}{2})$$

Mahia ngā tāpiri i roto o -5+10+\left(5+10\right)i.

$$Re(\frac{5+15i}{2})$$

Whakawehea te 5+15i ki te 2, kia riro ko \frac{5}{2}+\frac{15}{2}i.

$$Re(\frac{5}{2}+\frac{15}{2}i)$$

Ko te wāhi tūturu o \frac{5}{2}+\frac{15}{2}i ko \frac{5}{2}.

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