evaluate: i+i²+i³+i⁴
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Answer:
The Question is under complex numbers
Step-by-step explanation:
In conplex numbers
1² = -1
1³ = (1²) × 1 = -i X
¡4 = (₁²) × (₁²) = -1 × -1 = 1
By substitution i+₁²+₁³+i4 = i + (−1) + (-i) + (1)
=i-1-1+1 = 0
Final answer is Zero(0
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