Currently, #-sqrt3 + i# is written in standard/rectangular form, or #a + bi#. To convert it into polar form, we must write it in the format #r(costheta + isintheta)#, where #r# is the radius.
The radius is like the hypotenuse in a right triangle. You solve it like this:
#r = sqrt(a^2 + b^2)#
In this case, we know that #a = -sqrt3# and #b = 1#, so we can substitute them into the formula:
#r = sqrt((-sqrt3)^2 + (1)^2)#
Now simplify:
#r = sqrt(3 + 1)#
#r = sqrt4#
#r = 2#
#--------------------#
To solve for #theta#, we use #theta = tan^-1(b/a)#.
Let's substitute our values into the formula again:
#theta = tan^-1(1/-sqrt3)#
#theta = tan^-1(-1/sqrt3)#
#theta = (5pi)/6# or #(11pi)/6#
However, to determine the angle, we need to look at the graph of our expression #-sqrt3 + i#
Here's the graph for it. The real axis is the #x#-axis and the imaginary axis is the #y#-axis. In the graph, the #color(red)("red dot")# is at #-sqrt3 + i#.
As we can see, the complex number is in the 2nd quadrant. Looking back at our angle, we know that #theta# is either #(5pi)/6# or #(11pi)/6#. Since #(5pi)/6# is in the second quadrant, we know that #(5pi)/6# is #theta#.
#--------------------#
Finally, we write this in polar form.
We know that #r = 2# and #theta = (5pi)/6#, so let's substitute it into polar form:
#quadquadquadr(costheta + isintheta)#
So the final answer is:
#=> 2[cos((5pi)/6) + isin((5pi)/6)]#
*The #[]# brackets means the same as parenthesis, it's just for easier read.
#--------------------#
Instead of writing the long #2[cos((5pi)/6) + isin((5pi)/6)]#, you can also write it as a shortcut, as shown here:
#2cis(5pi)/6#
Where "#cis#" refers to "#cosisin#"
However, this is only if you teacher allows it ! If not, keep it as #2[cos((5pi)/6) + isin((5pi)/6)]#!
Hope this helps!
