Whakaoti i te z=i^3-1+2i/2-i

$$z = i ^ { 3 } - \frac { 1 + 2 i } { 2 - i }$$

Whakaoti mō z

$$z=-2i$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tātaihia te i mā te pū o 3, kia riro ko -i.

$$z=-i-\frac{1+2i}{2-i}$$

Me whakarea te taurunga me te tauraro o \frac{1+2i}{2-i} ki te haumi hiato o te tauraro, 2+i.

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

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

$$z=-i-\frac{5i}{5}$$

Whakawehea te 5i ki te 5, kia riro ko i.

$$z=-i-i$$

Tangohia te i i te -i, ka -2i.

$$z=-2i$$

Tautapa z

$$z≔-2i$$