Answer:
z
2
+z+1=0⇒z=ω,ω
∴(z+
1
)
+(z
+
+......(z
6
=(ω+ω
+(ω+ω
+(ω
3
ω
4(ω+ω
+2(ω
=4(1)+2(2
)=4+8=12.
[tex]z ^3 [/tex]
Step-by-step explanation:
z×(z^2)^1
=> z×(z^2×1)
=> z× (z^2)
=> z^3
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Answers & Comments
Answer:
z
2
+z+1=0⇒z=ω,ω
2
∴(z+
z
1
)
2
+(z
2
+
z
2
1
)
2
+......(z
6
+
z
6
1
)
2
=(ω+ω
2
)
2
+(ω+ω
2
)
2
+(ω
3
+
ω
3
1
)
2
+(ω+ω
2
)
2
+(ω+ω
2
)
2
+(ω
6
+
ω
6
1
)
2
4(ω+ω
2
)
2
+2(ω
3
+
ω
3
1
)
2
=4(1)+2(2
2
)=4+8=12.
Verified answer
Answer:
[tex]z ^3 [/tex]
Step-by-step explanation:
z×(z^2)^1
=> z×(z^2×1)
=> z× (z^2)
=> z^3