Iota and Theta Day Problem

${ z }^{ 5 }=4-4i$

Writing this in polar form we get :

${ z }^{ 5 }={ 2 }^{ (5/2) }{ e }^{ \frac { -i\pi }{ 4 } }$

Applying de moviere's formula we get :

$z=\sqrt { 2 } { e }^{ \frac { -i\pi }{ 20 } +\frac { 2k\pi }{ 5 } }=\sqrt { 2 } { e }^{ \frac { (8k-1)i\pi }{ 20 } }$

Comparing with the given equation we get :

${ K }_{ j }=(8k-1)$ where $k\epsilon I$

Put the values of k to get :

${ K }_{ j }=7,15,23,31,39$

Sum of all the ${K}_{j} is =7+15+23+31+39=\boxed{115}$

${z}^{5}=4-4i=4\sqrt{2}{e}^{i(2\pi k - \frac{\pi}{4})}$ , through writing the complex number in exponential form.

The expression $2\pi k - \frac{\pi}{4}$ comes from the argument being $-\frac{\pi}{4}$ and adding any multiple of $2\pi$ to the argument keeps the value the same.

$\therefore z=\sqrt{2}{e}^{\frac{i\pi}{20}(8k-1)}$

Where the expression $8k-1=K_j$

Hence plugging in suitable integer values of $k$ , ( $1,2,3,4,5$ ), it can be found that the values of $K_j$ in the range $0<K_j<40$ are $7,15,23,31,39$ .

Summing these results in $\boxed{115}$

There is nothing wrong with Ronak Agarwal's approach, except he should have specified the difference between De Moivre's theorem and De Moivre's Theorem for finding roots: $z={r}^{1/n}cis\left(\frac{\theta+2\pi k}{n}\right)$ where $k$ ranges from 0 to $n-1$ .

The modulus is indeed $\sqrt{2}$ because we have $r= {\left(\sqrt{32}\right)}^{1/5}$ . We then solve for the arguments: $\frac{{tan}^{-1}(-4/4) + 2\pi k}{5}$ .

Hence we get radian measures of $\frac{7}{20},\frac{15}{20},\frac{15}{20},\frac{23}{20},\frac{31}{20}$ . Adding the numerators, we get 115.