Question 981148
{{{2-2i}}} is in the form {{{a+bi}}} where {{{a = 2}}} and {{{b = -2}}}


Trig form is {{{r*(cos(theta) + i*sin(theta))}}}. We will use the following formulas:


{{{r = sqrt(a^2 + b^2)}}}
{{{theta = arctan(b/a)}}}
-------------------------------------------------------
Compute the value of r
{{{r = sqrt(a^2 + b^2)}}}
{{{r = sqrt(2^2 + (-2)^2)}}}
{{{r = sqrt(4+4)}}}
{{{r = sqrt(8)}}}
{{{r = sqrt(4*2)}}}
{{{r = sqrt(4)*sqrt(2)}}}
{{{r = 2*sqrt(2)}}}
-------------------------------------------------------
Now find theta (use a calculator or unit circle).
{{{theta = arctan(b/a)}}}
{{{theta = arctan(-2/2)}}}
{{{theta = arctan(-1)}}}
{{{theta = -pi/4}}}
{{{theta = -pi/4+2pi}}} Adding on 2pi to get the angle between 0 and 2pi
{{{theta = 7pi/4}}}
-------------------------------------------------------
So the trig form is
{{{r*(cos(theta) + i*sin(theta))=2*sqrt(2)*(cos(7pi/4) + i*sin(7pi/4))}}}
-------------------------------------------------------
Final Answer: {{{2*sqrt(2)*(cos(7pi/4) + i*sin(7pi/4))}}}


Note: this is the radian form