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junior college 2 | H2 Maths
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Help jake! Anyone can contribute an answer, even non-tutors.
complex numbers
Date Posted: 8 months ago
Views: 164
Converting to exponential form you get
z* = ((1 + a²)^(1.5) exp(iarctan(a)))/(2 exp(-iπ/2))
Taking the magnitude
(1 + a²)^(1.5)/2 = 4
(1 + a²)^(1.5) = 8
1 + a² = 4
a = sqrt(3)
Substitute a back in z*
z* = 4 exp(i5π/6)
z = 4 exp(-i5π/6)
z* = ((1 + a²)^(1.5) exp(iarctan(a)))/(2 exp(-iπ/2))
Taking the magnitude
(1 + a²)^(1.5)/2 = 4
(1 + a²)^(1.5) = 8
1 + a² = 4
a = sqrt(3)
Substitute a back in z*
z* = 4 exp(i5π/6)
z = 4 exp(-i5π/6)
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