One of the solutions of equation z cube = w is 21 = -5/3 + 51. What is one of other two solutions if w is a complex number?
Answers
Answer:
HII BRAINLY USERS YOUR SOLUTION
z
1
+z
2
∣
2
=∣z
1
∣
2
+∣z
2
∣
2
⇒(z
1
+z
2
)(
z
1
ˉ
+
z
2
ˉ
)=∣z
1
∣
2
+∣z
2
∣
2
⇒z
1
z
2
ˉ
+
z
1
ˉ
z
2
=0
⇒z
1
z
2
ˉ
+
z
1
z
2
ˉ
ˉ
=0
Therefore, z
1
z
2
ˉ
is purely imaginary
⇒
z
2
z
1
+
z
2
ˉ
z
1
ˉ
=0
also
z
2
z
1
is purely imaginary.
⇒arg(
z
2
z
1
)=
2
π
and hence, O,z
1
,z
2
are vertices of right triangle.
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Answer:
Whenever you’re dealing with complex roots, remember complex roots of unity. The answer comes in a similar way. So immediately convert into polar coordinates.
z1=−53–√+5iz1=100⋅e−π6iz1=−53+5iz1=100⋅e−π6i
Since that is one of the solutions, all of the solutions can be expressed with the formula
z=100⋅e(−π6+k⋅23π)iz=100⋅e(−π6+k⋅23π)i, where k=0,1,2k=0,1,2.
Step-by-step explanation:
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