Jtdylktuy

Find trigonometric form of complex number $frac{1}{2}-frac{1}{sqrt{2}}i$.

I assume that this is related to trigonometric form of complex number

$$1 pm cos(alpha) pm isin(alpha)$$ or similiar.

Is it possible to find such form without taking arctangent?

Just do it.

$|frac 12 - frac 1{sqrt 2} i| = sqrt {frac 12^2 + frac 12} = frac {sqrt 3} 2$

So $frac 12 - frac 1{sqrt 2} i = frac {sqrt 3}2 (frac 1{sqrt 3} - frac {sqrt 2}{sqrt 3}i)$

$frac {1}{sqrt 3}^2 + (-frac {sqrt 2}{sqrt 3})^2 = 1$ so there exists a unique $theta$ ($0 le theta < 2pi$) so that $cos theta = frac 1{sqrt 3}$ and $sin theta = -frac {sqrt 2}{sqrt 3}$.

In that case $tan theta = frac {sin theta}{cos theta} = frac {-frac {sqrt 2}{sqrt 3}}{frac 1{sqrt 3}} = -{sqrt 2}$.

So $theta$ would be $arctan {-sqrt 2}$ with some possible linear transformation to put it in the proper (4th) quadrant. As $arctan$ returns values from $-frac pi 2$ to $ frac pi 2$ (the 4th and 1st quadrant) we are good.

$theta = arctan{-sqrt 2} = -0.304087... pi$. Which so far as I can tell has no rational interpretation.

So $ frac 12 - frac 1{sqrt 2} i = frac {sqrt 3}2(cos arctan{-sqrt 2} + sin arctan{-sqrt 2} i) = frac {sqrt 3}2e^{iarctan{-sqrt 2}}$

You should write in form $$ z= |z|(cos alpha +isin alpha)$$

The modulus $$|z| = sqrt{{1over 4}+{1over 2}} = {sqrt{3}over2}$$

Since $sin alpha = -{sqrt{2}over sqrt{3}}$ and $cos alpha = -{1over 2sqrt{3}}$ and you will get an $alpha$ by solving this system.

HINT

You have
$$
re^{it} = rleft(cos t + isin tright),
$$
and note that
$$
r^2 = left(frac12right)^2+left(frac1{sqrt2}right)^2 = frac34,
$$
can you find $t$?

In general any complex number $z=x+iy$ can be represented as $re^{itheta}$ where begin{equation} r^2 = |z|^2 = x^2+y^2end{equation} and begin{equation} theta = tan^{-1} Bigl(frac{y}{x}Bigr)end{equation}

Here, begin{equation}
r = sqrt{frac{1}{4}+frac{1}{2}} = frac{sqrt{3}}{2}end{equation}
begin{equation}
theta = tan^{-1}Bigl(frac{-frac{1}{sqrt{2}}}{frac{1}{2}}Bigr) = tan^{-1}(-sqrt{2})
end{equation}
Here, since $ costheta > 0 $ and $ sintheta <0, theta $ is in fourth quadrant. Thus, $ frac{3π}{2}lethetale2π $ or simply,
$theta = -tan^{-1} (sqrt{2})$

Now assume that begin{equation}
tan{x} = u
end{equation}
Thus, begin{equation}
cos = sqrt{frac{1}{1+u^2}}
end{equation}
Thus,
begin{equation}
x = tan^{-1} (u) = cos^{-1}Bigl(sqrt{frac{1}{1+u^2}}Bigr)
end{equation}
Thus,
begin{equation}
cos(tan^{-1} (u)) = frac{1}{sqrt{1+u^2}}
end{equation}
Thus,
begin{equation} cos(tan^{-1}(-sqrt{2})) = frac{1}{sqrt{3}}end{equation}
Thus,begin{equation} sin{theta} =- sqrt{frac{2}{3}}end{equation}