Whakaoti i te (1+i)z+5=2-3i

$$( 1 + i ) z + 5 = 2 - 3 i$$

Whakaoti mō z

$$z=-3$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tangohia te 5 mai i ngā taha e rua.

$$\left(1+i\right)z=2-3i-5$$

Tangohia te 5 i te 2-3i mā te tango i ngā wāhi tūturu me ngā wāhi pohewa hāngai.

$$\left(1+i\right)z=2-5-3i$$

Tangohia te 5 i te 2, ka -3.

$$\left(1+i\right)z=-3-3i$$

Whakawehea ngā taha e rua ki te 1+i.

$$z=\frac{-3-3i}{1+i}$$

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

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

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}.

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

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

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

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

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

Hei tōna tikanga, ko te i^{2} ko -1.

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

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

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

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

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

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

$$z=\frac{-6}{2}$$

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

$$z=-3$$