Question 420873
{{{(16(cos(300)+i*sin(300)))/(8(cos(60)+i*sin(60)))}}}
Dividing complex numbers written in polar form is fairly simple because there is a formula that makes it easy. For two complex numbers:
{{{z[1] = r[1](cos(x[1]) + i*sin(x[1]))}}}
and 
{{{z[2] = r[2](cos(x[2]) + i*sin(x[2]))}}}
we can divide them using the formula:
{{{z[1]/z[2] = (r[1]/r[2])(cos(x[1]-x[2]) + i*sin(x[1]-x[2]))}}} (as long as {{{z[2]}}} is not zero).<br>
Using this formula to divide your complex numbers we get:
{{{(16/8)(cos(300-60) + i*sin(300-60))}}}
which simplifies to
{{{(2)(cos(240) + i*sin(240))}}}
which is the answer in polar form. For standard form, a+bi, we replace the cos and sin with their values. 240 is a special angle so we don't need a calculator:
{{{(2)((-1/2) + i*(-sqrt(3)/2))}}}
Distributing the 2 we get:
{{{-1 + i*(-sqrt(3))}}}
or
{{{-1 + (-sqrt(3))i}}}