if z² = -3 + 4i , then is it true that z= ±( 1+2i) ?
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Answer:
z
1
2
−z
2
2
∣=∣
z
4
2
+
z
4
2
−2
z
1
z
2
∣
⇒∣z
1
−z
2
∣∣z
1
+z
2
∣=∣
z
1
−
z
2
∣
2
⇒∣z
1
+z
2
∣=∣
z
1
−
z
2
∣
∣z
1
+z
2
∣=∣z
1
−z
2
∣
⇒
∣
∣
∣
∣
∣
z
2
z
1
+1
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
z
2
z
1
−1
∣
∣
∣
∣
∣
⇒
z
2
z
1
lies on ⊥ bisector of 1 and -1
⇒
z
2
z
1
lies on imaginary axis
⇒
z
2
z
1
is purly imaginary
⇒arg(
z
2
z
1
)=±
2
π
∣arg(z
1
)−arg(z
2
)∣=
2
π
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