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Question: If \({e^{x + iy}} = \alpha + i\beta \), then \(x + iy\) is called logarithm of \(\alpha + i\beta \) ...

If ex+iy=α+iβ{e^{x + iy}} = \alpha + i\beta , then x+iyx + iy is called logarithm of α+iβ\alpha + i\beta to the base ee: log⁡e(x+iy)=log⁡e(reiθ)=log⁡er+iθ{\log _e}\left( {x + iy} \right) = {\log _e}\left( {r{e^{i\theta }}} \right) = {\log _e}r + i\theta
Where, rr is modulus value of x+iyx + iy and θ\theta be the argument of x+iyx + iy if i(α+iβ)=α+iβ{i^{\left( {\alpha + i\beta } \right)}} = \alpha + i\beta , then α2+β2{\alpha ^2} + {\beta ^2} equals
A. e(2n+1)πn2{e^{\left( {2n + 1} \right)\dfrac{{\pi n}}{2}}}
B. e−πβ{e^{ - \pi \beta }}
C. e(4n+1)αβ{e^{\left( {4n + 1} \right)\alpha \beta }}
D. None  of  these{\text{None}}\;{\text{of}}\;{\text{these}}

Explanation

Solution

In this question we will use the property of complex numbers and use logarithm to simplify the equations and apply the condition as given in the question. Compare the imaginary parts separately to get the accurate answer.

Complete step by step solution:
As we know that a complex number can be expressed as x+iyx + iy, here xx is the real term and yy is the imaginary term.
According to the question it is given that,
i(α+iβ)=α+iβ{i^{\left( {\alpha + i\beta } \right)}} = \alpha + i\beta ………...….. (1)
Take log on both sides in the equation (1).
(α+iβ)log⁡i=log⁡(α+iβ)\left( {\alpha + i\beta } \right)\log i = \log \left( {\alpha + i\beta } \right)……..……. (2)
Apply the condition of log⁡e(x+iy)=log⁡e(reiθ){\log _e}\left( {x + iy} \right) = {\log _e}\left( {r{e^{i\theta }}} \right) in the equation (2).
Here, rr is the modulus value or magnitude value of x+iyx + iy and θ\theta be the argument of x+iyx + iy.
So,
log⁡(α+iβ)=(α+iβ)[iπ2+2πi],nεI\log \left( {\alpha + i\beta } \right) = \left( {\alpha + i\beta } \right)\left[ {i\dfrac{\pi }{2} + 2\pi i} \right],n\varepsilon I……………...(3)
Now, apply the formula of log⁡(α+iβ)\log \left( {\alpha + i\beta } \right)n equation (3) and multiply by ii on both sides.
12log⁡(α2+β2)+itan⁡−1βα=i(α+iβ)(4n+1)π2 12log⁡(α2+β2)+itan⁡−1βα=iα(4n+1)π2−β(4n+1)π2 \dfrac{1}{2}\log \left( {{\alpha ^2} + {\beta ^2}} \right) + i{\tan ^{ - 1}}\dfrac{\beta }{\alpha } = i\left( {\alpha + i\beta } \right)\left( {4n + 1} \right)\dfrac{\pi }{2} \\\ \dfrac{1}{2}\log \left( {{\alpha ^2} + {\beta ^2}} \right) + i{\tan ^{ - 1}}\dfrac{\beta }{\alpha } = i\alpha \left( {4n + 1} \right)\dfrac{\pi }{2} - \beta \left( {4n + 1} \right)\dfrac{\pi }{2} .........……..(4)
Compare the imaginary parts in the equation (4) and it is written as,
$
\dfrac{1}{2}\log \left( {{\alpha ^2} + {\beta ^2}} \right) = - \dfrac{\beta }{2}\left( {4n + 1} \right) \\

Therefore, Therefore,{\alpha ^2} + {\beta ^2} = {e^{ - \left( {4n + 1} \right)\pi \beta }}$

Hence, the correct option is D.

Note: As we know that the complex numbers are the combination of real and imaginary parts which can be expressed as x+iyx + iy. Complex numbers enlarge the concept of 1−D1 - D number line to 2−D2 - D complex plane. The concept of complex numbers is widely used in the field of mathematics and physics. If the real part of the complex number is zero then the complex number is purely imaginary and if the imaginary part of the complex number is zero then the complex number is purely real.