Aromātai
\frac{4}{25}+\frac{3}{25}i=0.16+0.12i
Wāhi Tūturu
\frac{4}{25} = 0.16
Tohaina
Kua tāruatia ki te papatopenga
\frac{1\left(4+3i\right)}{\left(4-3i\right)\left(4+3i\right)}
Whakareatia te taurunga me te tauraro ki te haumi hiato o te tauraro, 4+3i.
\frac{1\left(4+3i\right)}{4^{2}-3^{2}i^{2}}
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{1\left(4+3i\right)}{25}
Hei tōna tikanga, ko te i^{2} ko -1. Tātaitia te tauraro.
\frac{4+3i}{25}
Whakareatia te 1 ki te 4+3i, ka 4+3i.
\frac{4}{25}+\frac{3}{25}i
Whakawehea te 4+3i ki te 25, kia riro ko \frac{4}{25}+\frac{3}{25}i.
Re(\frac{1\left(4+3i\right)}{\left(4-3i\right)\left(4+3i\right)})
Me whakarea te taurunga me te tauraro o \frac{1}{4-3i} ki te haumi hiato o te tauraro, 4+3i.
Re(\frac{1\left(4+3i\right)}{4^{2}-3^{2}i^{2}})
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{1\left(4+3i\right)}{25})
Hei tōna tikanga, ko te i^{2} ko -1. Tātaitia te tauraro.
Re(\frac{4+3i}{25})
Whakareatia te 1 ki te 4+3i, ka 4+3i.
Re(\frac{4}{25}+\frac{3}{25}i)
Whakawehea te 4+3i ki te 25, kia riro ko \frac{4}{25}+\frac{3}{25}i.
\frac{4}{25}
Ko te wāhi tūturu o \frac{4}{25}+\frac{3}{25}i ko \frac{4}{25}.
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