In Fig. 7, find ZABC + ZBCD + ZCDE + ZDEF + ZEFA + ZFAB.
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Answer:
Given,
z
1/3
=.p+iq
or,z=(p+iq)
3
or, x−iy=(p
3
−3pq
2
)+i(3p
2
q−q
3
)
Comparing both sides we get,
x=p
3
−3pq
2
and y=−3p
2
q+q
3
.
Now,
p
x
+
q
y
=p
2
−3q
2
−3p
2
+q
2
=−2(p
2
+q
2
)......(1).
Now,
p
2
+q
2
p
x
+
q
y
=−2. [ Using (1)]
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