Whakaoti i te z=20t/3-i-(5-3i)*(2+3i)^2+(1+i)^5

$$z = \frac { 20 t } { 3 - i } - ( 5 - 3 i ) \cdot ( 2 + 3 i ) ^ { 2 } + ( 1 + i ) ^ { 5 }$$

Whakaoti mō t

$$t=\left(\frac{3}{20}-\frac{1}{20}i\right)z+\left(\frac{31}{5}+\frac{111}{10}i\right)$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakawehea te 20t ki te 3-i, kia riro ko \left(6+2i\right)t.

$$z=\left(6+2i\right)t-\left(5-3i\right)\left(2+3i\right)^{2}+\left(1+i\right)^{5}$$

Tātaihia te 2+3i mā te pū o 2, kia riro ko -5+12i.

$$z=\left(6+2i\right)t-\left(5-3i\right)\left(-5+12i\right)+\left(1+i\right)^{5}$$

Whakareatia te 5-3i ki te -5+12i, ka 11+75i.

$$z=\left(6+2i\right)t-\left(11+75i\right)+\left(1+i\right)^{5}$$

Tātaihia te 1+i mā te pū o 5, kia riro ko -4-4i.

$$z=\left(6+2i\right)t-\left(11+75i\right)+\left(-4-4i\right)$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$\left(6+2i\right)t-\left(11+75i\right)+\left(-4-4i\right)=z$$

Me tāpiri te 4+4i ki ngā taha e rua.

$$\left(6+2i\right)t-\left(11+75i\right)=z+\left(4+4i\right)$$

Me tāpiri te 11+75i ki ngā taha e rua.

$$\left(6+2i\right)t=z+\left(4+4i\right)+\left(11+75i\right)$$

Mahia ngā tāpiri i roto o 4+4i+\left(11+75i\right).

$$\left(6+2i\right)t=z+15+79i$$

He hanga arowhānui tō te whārite.

$$\left(6+2i\right)t=z+\left(15+79i\right)$$

Whakawehea ngā taha e rua ki te 6+2i.

$$\frac{\left(6+2i\right)t}{6+2i}=\frac{z+\left(15+79i\right)}{6+2i}$$

Mā te whakawehe ki te 6+2i ka wetekia te whakareanga ki te 6+2i.

$$t=\frac{z+\left(15+79i\right)}{6+2i}$$

Whakawehe z+\left(15+79i\right) ki te 6+2i.

$$t=\left(\frac{3}{20}-\frac{1}{20}i\right)z+\left(\frac{31}{5}+\frac{111}{10}i\right)$$

Whakaoti mō z

$$z=\left(6+2i\right)t+\left(-15-79i\right)$$