Squared complex numbers. Square (4+3i)^2.
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Tushar Chandra
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We have to find the square of (4+3i)
(4+3i)^2
=> (4+3i)(4+3i)
=> 4*4 + 4*3*i + 4*3*i + 3*i*3*i
=> 16 + 12i + 12i + 9i^2
i^2 = -1
=> 16 + 24i - 9
=> 7 + 24i
The required square is 7 + 24i
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giorgiana1976 | Student
To square the complex number, we'll apply the formula:
(a+b)^2 = a^2 + 2ab + b^2
We'll put a = 4 and b = 3i and we'll expand the binomial:
(4+3i)^2 = 4^2 + 2*4*3i + (3i)^2
(4+3i)^2 = 16 + 24i + 9i^2
We'll replace i^2 by -1:
(4+3i)^2 = 16 + 24i - 9
We'll combine the real parts:
(4+3i)^2 = 7 + 24i
The complex number raised to square is: (4+3i)^2 = 7 + 24i.
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