Question

Consider the expression $z=r{{e}^{i\theta }}$ , then the value of $\left| {{e}^{iz}} \right|$ is equal to
(a) ${{e}^{-r\sin \theta }}$
(b) $r{{e}^{-r\sin \theta }}$
(c) ${{e}^{-r\cos \theta }}$
(d) $r{{e}^{-r\cos \theta }}$

Answer 1

Hint: Use the Euler’s formula of sin, cos of a particular angle. Here the left-hand side can also be written as $Cisx$ it’s our wish as both are same
${{e}^{ix}}=\cos x+i\sin x$

Complete step-by-step solution -
Definition of i, can be written as:
The solution of the equation: ${{x}^{2}}+1=0$ is i. i is an imaginary number. Any number which has an imaginary number in its representation is called a complex number.
Definition of a complex equation, can be written as: An equation containing complex numbers in it, is called a complex equation.
It is possible to have a real root for complex equations.
Example: $\left( 1+i \right)x+\left( 1+i \right)=0,x= -1$ is the root of the equation.
Given expression in the question of the variable z is:
\[z=r{{e}^{i\theta }}\]
For our easy representation we take $\theta $ as q in our solution
$z=r{{e}^{iq}}$
By Euler’s formula of sin, cos of an angle q is:
${{e}^{iq}}=\cos q+i\sin q$
By substituting this into out term z, we turn z into:
$z=r\left( \cos q+i\sin q \right)$
We need the term iz for our required expression in question.
So, by multiplying with i on both sides, we get:
\[\begin{align}
  & iz=ri\left( \cos q+i\sin q \right) \\
 & iz=r\left( -\sin q+i\cos q \right) \\
\end{align}\]
The required expression of modulus of e term is:
$\left| {{e}^{iz}} \right|$
We know the value of iz by the above equations
By substituting that value here in our required expression, we get:
$\left| {{e}^{iz}} \right|=\left| {{e}^{r\left( -\sin q+i\cos q \right)}} \right|$
By basic exponential knowledge, we have a formula on e:
${{e}^{a+b}}={{e}^{a}}.{{e}^{b}}$
Simplifying our required expression of modulus, we get:
$\left| {{e}^{iz}} \right|=\left| {{e}^{-r\sin q+ri\cos q}} \right|$
By using above equation, we get:
$\left| {{e}^{iz}} \right|=\left| {{e}^{-r\sin q}} \right|.\left| {{e}^{ri\cos q}} \right|$
We know: $\left| {{e}^{ik}} \right|=1\text{ }\left( \text{as }{{\cos }^{2}}x+{{\sin }^{2}}x=1 \right)$
By using above equation, we get:
\[\left| {{e}^{iz}} \right|=\left| {{e}^{-r\sin q}} \right|\] = \[{{e}^{-r\sin q}}\] is always positive
\[\left| {{e}^{iz}} \right|={{e}^{-r\sin q}}\]
Option (a) is correct

Note: $\left| {{e}^{ik}} \right|=1$ as, ${{e}^{ik}}=\sin k+i\operatorname{cosk}\Rightarrow \left| {{e}^{ik}} \right|=\sqrt{{{\sin }^{2}}+{{\cos }^{2}}k}=1$
This is an important proof used in the solution. Here it is need to be careful while using the modulus of $ \left| {{e}^{iz}} \right|