Question

If ${e^{x + iy}} = \alpha + i\beta $, then $x + iy$ is called logarithm of $\alpha + i\beta $ to the base $e$: ${\log _e}\left( {x + iy} \right) = {\log _e}\left( {r{e^{i\theta }}} \right) = {\log _e}r + i\theta $
Where, $r$ is modulus value of $x + iy$ and $\theta $ be the argument of $x + iy$ if ${i^{\left( {\alpha + i\beta } \right)}} = \alpha + i\beta $, then ${\alpha ^2} + {\beta ^2}$ equals
A. ${e^{\left( {2n + 1} \right)\dfrac{{\pi n}}{2}}}$
B. ${e^{ - \pi \beta }}$
C. ${e^{\left( {4n + 1} \right)\alpha \beta }}$
D. ${\text{None}}\;{\text{of}}\;{\text{these}}$

Answer 1

Hint: 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 $x + iy$, here $x$ is the real term and $y$ is the imaginary term.
According to the question it is given that,
${i^{\left( {\alpha + i\beta } \right)}} = \alpha + i\beta $ ………...….. (1)
Take log on both sides in the equation (1).
$\left( {\alpha + i\beta } \right)\log i = \log \left( {\alpha + i\beta } \right)$……..……. (2)
Apply the condition of ${\log _e}\left( {x + iy} \right) = {\log _e}\left( {r{e^{i\theta }}} \right)$ in the equation (2).
Here, $r$ is the modulus value or magnitude value of $x + iy$ and $\theta $ be the argument of $x + iy$.
So,
$\log \left( {\alpha + i\beta } \right) = \left( {\alpha + i\beta } \right)\left[ {i\dfrac{\pi }{2} + 2\pi i} \right],n\varepsilon I$……………...(3)
Now, apply the formula of $\log \left( {\alpha + i\beta } \right)$n equation (3) and multiply by $i$ on both sides.
$
  \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, ${\alpha ^2} + {\beta ^2} = {e^{ - \left( {4n + 1} \right)\pi \beta }}$

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 $x + iy$. Complex numbers enlarge the concept of $1 - D$ number line to $2 - D$ 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.