Question

separate into real and imaginary parts of tanh(1+i)​

Answer 1

Answer:

Let (1 + i) = z, then

tanh(z)={e^(z)-e^(-z)}/{e^(z)+e^(-z)}

= Nr/Dr . Now e^(z) = e^(1+i) = e.e^(i) and e^(-z) = e^(-1-i) = e^(-1).e^(-i) so that Nr/Dr

= {e.e^(i) -(1/e)e^(-i)}/{e.e^(i) + (1/e)e^(i)} = {e²e^(i) - e^(-i)}/{e²e^(i) + e^(-i)}

Multiplying the numerator (Nr) & denominator (Dr) by the complex conjugate of Dr i.e. by (e²e^(-i) + e^(i)), we get Dr = (e⁴+2e²cos2 + 1) and

Nr =(e⁴−1) + i (2e²sin2) . Therefore,

tanhz = (e⁴−1)/(e⁴+2e²cos2 +1)+i (2e²sin2)/(e⁴+2e²cos2 +1) etc.

Step-by-step explanation:

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