Question 816172
{{{z^4=iz}}}
Subtracting iz:
{{{z^4-iz=0}}}
Factor out z:
{{{z(z^3-i)=0}}}
From this we can see that z = 0 will be a root. The other (non-zero) roots will be solutions to {{{z^3-i=0}}}. In other words the other (non-zero) roots are the cube roots of i:<br>
{{{i^(1/3)}}}
In mod-arg form:
{{{(1cis(pi/2))^(1/3)}}}
To find these cube roots we will use DeMoivre's (spelling?) Theorem which tells us a power of a complex number in mod-arg form can be found by raising the mod to the power and multiplying the arg by the power:
{{{(1)^(1/3)cis((pi/2)*(1/3))}}}
which simplifies to:
{{{1cis(pi/6)}}}
This is the root, v, that you had already found. To find the other roots we just use co-terminal args:
{{{(1cis(pi/2+2pi))^(1/3)}}}
{{{(1cis(5pi/2))^(1/3)}}}
{{{(1)^(1/3)cis((5pi/2)*(1/3))}}}
{{{1cis(5pi/6)}}}<br>
{{{(1cis(pi/2+2pi+2pi))^(1/3)}}}
{{{(1cis(9pi/2))^(1/3)}}}
{{{(1)^(1/3)cis((9pi/2)*(1/3))}}}
{{{1cis(9pi/6)}}}
{{{1cis(3pi/2)}}}<br>