# Compute e^{z} in the form x + iy and \left | e^{z} \right | where z equals (a) 3 + \pi i (b) 1...

## Question:

Compute {eq}e^{z}{/eq} in the form {eq}x + iy{/eq} and {eq}\left | e^{z} \right |{/eq} where {eq}z{/eq} equals

(a) {eq}3 + \pi i {/eq}

(b) {eq}1 + i {/eq}

(c) {eq}\frac{3}{4\pi} i{/eq}

## Imaginary number:

An imaginary number is defined as the combination of the real number and the imaginary number. The imaginary number is denoted by i. The square of the imaginary unit is always as negative unity.

Given data:

• {eq}z = 3 + \pi i {/eq}
• {eq}z = 1 + i {/eq}
• {eq}z = \dfrac{3}{4}\pi i {/eq}

(a)

From the given data

{eq}z = 3 + \pi i {/eq}

The

{e^z}

} ]is expressed in the form of {eq}x + iy {/eq} is

{eq}\begin{align*} {e^z} &= {e^{3 + \pi i}}\\ &= {e^3} \cdot {e^{\pi i}}\cdots\cdots\rm{(I)} \end{align*} {/eq}

As we know that,

{eq}{e^{i\theta }} = \cos \theta + i\sin \theta {/eq}

The above formula substituting in equation (I) as,

{eq}\begin{align*} {e^z} &= {e^3} \cdot \left( {\cos \pi + i\sin \pi } \right)\\ &= {e^3}\left( { - 1 + i0} \right)\\ &= {e^3}\left( { - 1} \right)\\ &= - {e^{ - 3}} + i0 \end{align*} {/eq}

The expression for {eq}\left| {{e^z}} \right| {/eq} is

{eq}\begin{align*} \left| {{e^z}} \right| &= \sqrt {{x^2} + {y^2}} \\ &= \sqrt {{{\left( { - {e^{ - 3}}} \right)}^2} + {0^2}} \\ &= {e^3} \end{align*} {/eq}

Thus the value of {eq}\left| {{e^z}} \right| = {e^3} {/eq}

(b)

From the given data

{eq}z = 1 + i {/eq}

The {eq}{e^z} {/eq} is expressed in the form of {eq}x + iy {/eq}is

{eq}\begin{align*} {e^z} &= {e^{1 + i}}\\ &= e \cdot {e^i} \end{align*} (2) {/eq}

As we know that,

{eq}{e^{i\theta }} = \cos \theta + i\sin \theta {/eq}

The above formula substituting in equation (II) as,

{eq}{e^z} = e \cdot e\left( {\cos \left( 1 \right) + i\sin \left( 1 \right)} \right) {/eq}

The expression for {eq}\left| {{e^z}} \right| {/eq} is

{eq}\begin{align*} \left| {{e^z}} \right| &= e\sqrt {{{\cos }^2}\left( 1 \right) + {{\sin }^2}\left( 1 \right)} \\ &= e\sqrt 1 \\ &= e \end{align*} {/eq}

Thus the value of {eq}\left| {{e^z}} \right| = e {/eq}

(C)

From the given data

{eq}z = \dfrac{3}{4}\pi i {/eq}

The {eq}{e^z} {/eq} is expressed in the form of {eq}x + iy {/eq}is

{eq}\begin{align*} {e^z} &= {e^{\dfrac{3}{4}\pi i}}\\ &= \cos \left( {\dfrac{{3\pi }}{4}} \right) + i\sin \left( {\dfrac{{3\pi }}{4}} \right)\\ &= \cos \left( {\pi - \dfrac{\pi }{4}} \right) + i\sin \left( {\pi - \dfrac{\pi }{4}} \right)\\ &= - \cos \left( {\dfrac{\pi }{4}} \right) + i\sin \left( {\dfrac{\pi }{4}} \right)\\ &= \left( { - \dfrac{1}{{\sqrt 2 }} + i\dfrac{1}{{\sqrt 2 }}} \right) \end{align*} {/eq}

The expression for {eq}\left| {{e^z}} \right| {/eq} is

{eq}\begin{align*} \left| {{e^z}} \right| &= \sqrt {{{\left( { - \dfrac{1}{{\sqrt 2 }}} \right)}^2} + {{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2}} \\ &= \sqrt {\dfrac{1}{2} + \dfrac{1}{2}} \\ &= \sqrt 1 \\ &= 1 \end{align*} {/eq}

Thus the value of {eq}\left| {{e^z}} \right| = 1 {/eq}