Let a = 2 3 + lo g b ( 2 5 1 1 5 + 2 5 2 5 i ) + i π lo g b e and b a = z , where z is of the form A + B i . If ∣ z ∣ 2 + ℜ ( z ) + ℑ ( z ) = 1 1 2 , what is the value of b ?
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Isn't this with the assumption that b is real, which need not be the case?
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Isn't the logarithm function defined for only positive bases (excluding 1)?
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That is true only for the ' Real Logarithmic Function '. The complex analogue deals with all complex bases. Explicitly, one may define lo g a b = ln a ln b , { a , b ∈ C , ln a = 0 }
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Powering from base b in both sides:
b a = b 3 / 2 ⋅ ( 2 5 1 1 5 + i 2 5 2 5 ) ⋅ ( b l o g b e ) i π
z = b 3 / 2 ⋅ e a t a n ( 2 / 1 1 ) ⋅ e i π
Then:
∣ z ∣ 2 = b 3
R e ( z ) = − b 3 / 2 2 5 1 1 5
I m ( z ) = − b 3 / 2 2 5 2 5
Summing up and multiplyting both sides by 2 5 :
2 5 b 3 − 1 3 5 b 3 / 2 − 2 8 0 0 = 0
Solving for b 3 / 2 :
b 3 / 2 = 5 0 1 3 5 ± 2 3 7 5
b 3 / 2 = 5 5
b 3 = 1 2 5
b = 5