Question

If ${i^{{i^{i...........\infty }}}} = x + iy$ then ${x^2} + {y^2} = ?$

Answer 1

Hint : We are asked to find the value of ${x^2} + {y^2}$ . We can see that the power series tends to infinity, use the basic concept of infinity and take down the power use logarithm on both sides. You also need to use the concept of complex numbers for this problem.

Complete step-by-step answer :
Given, ${i^{{i^{i...........\infty }}}} = x + iy$
We are asked to find the value of ${x^2} + {y^2}$
Let $z = x + iy$
Now, we have ${i^{{i^{i...........\infty }}}} = z$
We can take ${i^z} = z$ as the power tends to be infinite one more term will not affect the expression.
Now we have,
${i^z} = z$ (i)
Taking logarithm on both sides we have,
$\log {i^z} = \log z$
$\Rightarrow z\log i = \log z$ (ii)
$i$ can be written as ${e^{i\dfrac{\pi }{2}}}$ as ${e^{i\dfrac{\pi }{2}}} = \cos \left( {\dfrac{\pi }{2}} \right) + i\sin \left( {\dfrac{\pi }{2}} \right) = i$ .
Substituting $i$ as ${e^{i\dfrac{\pi }{2}}}$ in equation (ii) we get,
$z\log {e^{i\dfrac{\pi }{2}}} = \log z$
$\Rightarrow iz\dfrac{\pi }{2} = \log z$ (iii)
We have $z = x + iy$ is a complex number and this can also we written as
$z = x + iy = r{e^{i\theta }}$ (iv)
where $r$ is the radial distance and $\theta$ is the angle between x and y components.
Substituting $z = r{e^{i\theta }}$ on R.H.S on equation (iii) we get

\Rightarrow iz\dfrac{\pi }{2} = \log r + \log {e^{i\theta }} $$ $$ \Rightarrow iz\dfrac{\pi }{2} = \log r + i\theta $$ (v) $$r$$ can also be written as $$r = \sqrt {{x^2} + {y^2}} $$ . Substituting this value of $$r$$ and $$z = x + iy$$ in equation (v) we get, $$i(x + iy)\dfrac{\pi }{2} = \log \sqrt {{x^2} + {y^2}} + i\theta \\\ \Rightarrow ix\dfrac{\pi }{2} - y\dfrac{\pi }{2} = \log \sqrt {{x^2} + {y^2}} + i\theta \\\ \Rightarrow - y\dfrac{\pi }{2} + ix\dfrac{\pi }{2} = \log \sqrt {{x^2} + {y^2}} + i\theta $$ Equating the real part of the above equation we get, $$\log \sqrt {{x^2} + {y^2}} = - y\dfrac{\pi }{2} \\\ \Rightarrow \log {\left( {{x^2} + {y^2}} \right)^{\dfrac{1}{2}}} = - y\dfrac{\pi }{2} $$ $$\Rightarrow \dfrac{1}{2}\log \left( {{x^2} + {y^2}} \right) = - y\dfrac{\pi }{2} \\\ \Rightarrow \log \left( {{x^2} + {y^2}} \right) = - y\pi \\\ \Rightarrow {x^2} + {y^2} = {e^{ - y\pi }} $$ Therefore, the value of $${x^2} + {y^2}$$ is $${e^{ - y\pi }}$$ . **So, the correct answer is “ $${e^{ - y\pi }}$$ ”.** **Note** : A complex number can be written as $$a + ib$$ where $$a$$ is the real part and $$ib$$ represents the complex part. We can classify a complex number based on the values of $$a$$ and $$b$$ .When the term $$b$$ is zero, we can say the number is purely real and when $$a$$ is zero, we can say the number is purely imaginary. One more important point one should remember is that every real number can be expressed in the form of a complex number but complex numbers can be expressed as real numbers.