Complex Cosine

By Euler's formula, ${ e }^{ iz }=\cos { (z) } +i\sin { (z) }$ .

$\Rightarrow { e }^{ -iz }=\cos { (z) } -i\sin { (z) } \\ \Rightarrow { e }^{ iz }+{ e }^{ -iz }=2\cos { (z) } \\ \Rightarrow \cos { (z)=\frac { { e }^{ iz }+{ e }^{ -iz } }{ 2 } . }$

Hence, $\frac { { e }^{ iz }+{ e }^{ -iz } }{ 2 } =4\\ \Rightarrow { e }^{ iz }+{ e }^{ -iz }=8\quad$

Multiplying through by ${ e }^{ iz }$ yields ${ e }^{ 2iz }-8{ e }^{ iz }+1=0$ , which is a quadratic in ${ e }^{ iz}$ . Solving this quadratic (by the quadratic formula or completing square) gives the two solutions $z=-i\log _{ e }{ (4\pm \sqrt { 15 } ) }$ .

Now, the expression ${ e }^{ \frac { z }{ i } }={ e }^{ -\log _{ e }{ (4\pm \sqrt { 15 } ) } }\\ \qquad ={ e }^{ \log _{ e }{ { (4\pm \sqrt { 15 } ) }^{ -1 } } }\\ \qquad =\frac { 1 }{ 4\pm \sqrt { 15 } } \\ \qquad =4\mp \sqrt { 15 }$ .

Hence, $a=4, b=15$ and $a+b=\boxed { 19 }.$

Let z=ir because it's given that it is pure imaginary i= $\sqrt { -1 }$

cos(ir)=4 , sin(ir)= $\pm \sqrt { 15 } i$ [ ${ cos }^{ 2 }x+{ sin }^{ 2 }x=1$ ]

$\Rightarrow$ cos(ir)+isin(ir)=4 $\mp \sqrt { 15 }$

$\Rightarrow$ ${ e }^{ i\ast (ir) }$ =4 $\mp \sqrt { 15 }$

$\Rightarrow$ ${ e }^{ -r }$ =4 $\mp \sqrt { 15 }$

Now, ${ e }^{ \frac { ir }{ i } }$ = ${ e }^{ r }$ = $\frac { 1 }{ 4\mp \sqrt { 15 } } =4\pm \sqrt { 15 }$

There are more solutions

$z=-i\log(4 \pm \sqrt{15})+ 2 \pi k$ for $k=0,\pm 1,\pm 2,....$