Converting to polar form

nacho

Active member
I started of with attempting to convert the numerator first

$| 1 + i | = \sqrt{1^2+i^2}$
$= \sqrt{1-1} = 0$ ? this is wrong obviously, i dont see why its $\sqrt{2}$

for the second part

$|\sqrt{3} - i|= \sqrt{3+1} = 2$

$x = r \cos\theta$ $y = r\sin\theta$

$x = 2\cos\theta$ $y=2\sin\theta$

then $\theta = \frac{\pi}{3} and \frac{\pi}{6}$

$= 2(\cos\frac{\pi}{3} + \sin\frac{\pi}{6} = 2cis(\frac{\pi}{3})$ I don't see why this is wrong either

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Ackbach

Indicium Physicus
Staff member
The magnitude of a complex number $a+bi$ is given by $|a+bi|= \sqrt{(a+bi)(a-bi)}$. That is, you multiply a number by its complex conjugate, and then you take the square root.

ZaidAlyafey

Well-known member
MHB Math Helper
$$\displaystyle 1+i = \sqrt{2} \left( \frac{1}{\sqrt{2}}+i \frac{1}{\sqrt{2}}\right) = \sqrt{2} \, \text{cis} \left( \frac{\pi}{4} \right)$$

$$\displaystyle \sqrt{3}-i = 2 \left(\frac{\sqrt{3}}{2}- i \frac{1}{2} \right) = 2 \text{cis}\left( \frac{-\pi}{6}\right)$$

nacho

Active member
$\cos\theta = \frac{1}{\sqrt{2}}$
Therefore
$\theta = \frac{\pi}{4}$
and $\sin\theta=\frac{-1}{\sqrt{2}}$
therefore
$\theta = \frac{\pi}{4} + \pi = \frac{5\pi}{4}$
What have i done wrong for the $\sin\theta$ part?

Prove It

Well-known member
MHB Math Helper
The range of \displaystyle \begin{align*} y = \arcsin{(x)} \end{align*} is \displaystyle \begin{align*} \left[ -\frac{\pi}{2} , \frac{\pi}{2} \right] \end{align*}.