Answer:
z
1
′
=z
2
3
=(1+2i)(1−3i)(2+4i)
⇒2(1−3i+2i−6i
)(1+2i)
⇒2(1−i+6)(1+2i) (∴i
=−1)
⇒2(7−i)(1+2i)⇒2(7+14i−i−2i
)=2(7+13i+2)=2(9+13i)
Therefore, z
=(18+26i),z
=(36+52i),z
=(−126−182i)
=18+26i
=2z
=−7z
∴z
,2z
,−7z
are collinear
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Answers & Comments
Answer:
z
1
′
=z
1
z
2
z
3
=(1+2i)(1−3i)(2+4i)
⇒2(1−3i+2i−6i
2
)(1+2i)
⇒2(1−i+6)(1+2i) (∴i
2
=−1)
⇒2(7−i)(1+2i)⇒2(7+14i−i−2i
2
)=2(7+13i+2)=2(9+13i)
Therefore, z
1
′
=(18+26i),z
2
′
=(36+52i),z
3
′
=(−126−182i)
z
1
′
=z
1
z
2
z
3
=18+26i
z
2
′
=2z
1
z
2
z
3
=2z
1
′
z
3
′
=−7z
1
′
z
2
′
z
3
′
=−7z
1
′
∴z
1
z
2
z
3
,2z
1
z
2
z
3
,−7z
1
′
z
2
′
z
3
′
are collinear