Currently, -sqrt3 + i−√3+i is written in standard/rectangular form, or a + bia+bi. To convert it into polar form, we must write it in the format r(costheta + isintheta)r(cosθ+isinθ), where rr is the radius.
The radius is like the hypotenuse in a right triangle. You solve it like this:
r = sqrt(a^2 + b^2)r=√a2+b2
In this case, we know that a = -sqrt3a=−√3 and b = 1b=1, so we can substitute them into the formula:
r = sqrt((-sqrt3)^2 + (1)^2)r=√(−√3)2+(1)2
Now simplify:
r = sqrt(3 + 1)r=√3+1
r = sqrt4r=√4
r = 2r=2
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To solve for thetaθ, we use theta = tan^-1(b/a)θ=tan−1(ba).
Let's substitute our values into the formula again:
theta = tan^-1(1/-sqrt3)θ=tan−1(1−√3)
theta = tan^-1(-1/sqrt3)θ=tan−1(−1√3)
theta = (5pi)/6θ=5π6 or (11pi)/611π6
However, to determine the angle, we need to look at the graph of our expression -sqrt3 + i−√3+i
Here's the graph for it. The real axis is the xx-axis and the imaginary axis is the yy-axis. In the graph, the color(red)("red dot")red dot is at -sqrt3 + i−√3+i.
![enter image source here]()
As we can see, the complex number is in the 2nd quadrant. Looking back at our angle, we know that thetaθ is either (5pi)/65π6 or (11pi)/611π6. Since (5pi)/65π6 is in the second quadrant, we know that (5pi)/65π6 is thetaθ.
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Finally, we write this in polar form.
We know that r = 2r=2 and theta = (5pi)/6θ=5π6, 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.
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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!
