Question 420876
When posting fractions on Algebra.com, please<ol><li>Type a "(" (left parenthesis)</li><li>Type the entire numerator</li><li>Type ")/(" (the right parenthesis closes the numerator, the "/" represents the division, and the left parenthesis opens the denominator)</li><li>Type the entire denominator</li><li>Type a ")" (to close the denominator)</li></ol>
This makes the actual fraction easy to understand. When you post fractions the way you did this one, it can be difficult to read.<br>
Putting arguments of function in parentheses helps, too. So your expression, posted this way, would be:
(3(cos(395)+isin(395)))/(9(cos(65)+isin(65)))<br>
{{{(3(cos(395)+i*sin(395)))/(9(cos(65)+i*sin(65)))}}}
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:
{{{(3/9)(cos(395-65) + i*sin(395-65))}}}
which simplifies to
{{{(1/3)(cos(330) + i*sin(330))}}}
which is the answer in polar form. For standard form, a+bi, we replace the cos and sin with their values. 330 is a special angle so we don't need a calculator:
{{{(1/3)(sqrt(3)/2 + i*(-1/2))}}}
Distributing the 1/3 we get:
{{{sqrt(3)/6 + i*(-1/6)}}}
or
{{{sqrt(3)/6 + (-1/6)i}}}