Whakaoti i te (2+i)*z-(3-2i)/2z=4+3i-(2-5i)z

$$( 2 + i ) \cdot z - \frac { ( 3 - 2 i ) } { 2 } z = 4 + 3 i - ( 2 - 5 i ) z$$

Whakaoti mō z

$$z=\frac{4}{61}+\frac{78}{61}i\approx 0.06557377+1.278688525i$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakawehea te 3-2i ki te 2, kia riro ko \frac{3}{2}-i.

$$\left(2+i\right)z-\left(\frac{3}{2}-i\right)z=4+3i-\left(2-5i\right)z$$

Pahekotia te \left(2+i\right)z me \left(-\frac{3}{2}+i\right)z, ka \left(\frac{1}{2}+2i\right)z.

$$\left(\frac{1}{2}+2i\right)z=4+3i-\left(2-5i\right)z$$

Me tāpiri te \left(2-5i\right)z ki ngā taha e rua.

$$\left(\frac{1}{2}+2i\right)z+\left(2-5i\right)z=4+3i$$

Pahekotia te \left(\frac{1}{2}+2i\right)z me \left(2-5i\right)z, ka \left(\frac{5}{2}-3i\right)z.

$$\left(\frac{5}{2}-3i\right)z=4+3i$$

Whakawehea ngā taha e rua ki te \frac{5}{2}-3i.

$$z=\frac{4+3i}{\frac{5}{2}-3i}$$

Me whakarea te taurunga me te tauraro o \frac{4+3i}{\frac{5}{2}-3i} ki te haumi hiato o te tauraro, \frac{5}{2}+3i.

$$z=\frac{\left(4+3i\right)\left(\frac{5}{2}+3i\right)}{\left(\frac{5}{2}-3i\right)\left(\frac{5}{2}+3i\right)}$$

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}.

$$z=\frac{\left(4+3i\right)\left(\frac{5}{2}+3i\right)}{\left(\frac{5}{2}\right)^{2}-3^{2}i^{2}}$$

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

$$z=\frac{\left(4+3i\right)\left(\frac{5}{2}+3i\right)}{\frac{61}{4}}$$

Me whakarea ngā tau matatini 4+3i me \frac{5}{2}+3i pēnā i te whakarea huarua.

$$z=\frac{4\times \frac{5}{2}+4\times \left(3i\right)+3i\times \frac{5}{2}+3\times 3i^{2}}{\frac{61}{4}}$$

Hei tōna tikanga, ko te i^{2} ko -1.

$$z=\frac{4\times \frac{5}{2}+4\times \left(3i\right)+3i\times \frac{5}{2}+3\times 3\left(-1\right)}{\frac{61}{4}}$$

Mahia ngā whakarea i roto o 4\times \frac{5}{2}+4\times \left(3i\right)+3i\times \frac{5}{2}+3\times 3\left(-1\right).

$$z=\frac{10+12i+\frac{15}{2}i-9}{\frac{61}{4}}$$

Whakakotahitia ngā wāhi tūturu me ngā wāhi pōhewa ki 10+12i+\frac{15}{2}i-9.

$$z=\frac{10-9+\left(12+\frac{15}{2}\right)i}{\frac{61}{4}}$$

Mahia ngā tāpiri i roto o 10-9+\left(12+\frac{15}{2}\right)i.

$$z=\frac{1+\frac{39}{2}i}{\frac{61}{4}}$$

Whakawehea te 1+\frac{39}{2}i ki te \frac{61}{4}, kia riro ko \frac{4}{61}+\frac{78}{61}i.

$$z=\frac{4}{61}+\frac{78}{61}i$$