Question

Please help:
Write in polar form 1-sqrt(3)*i

Answer 1

\[1-\sqrt{3}i\]

Answer 2

\(\large \tt \begin{align} \color{black}{z=a+bi\\~\\
r=|z|=\sqrt{a^2+b^2}\\~\\
\alpha=tan^{-1}(\dfrac{b}{a})\\~\\
trig form \\~\\
z=r(cos\alpha+isin\alpha)\\~\\
} \end{align}\)

Answer 3

But is there a way to simplify
\[z=2(\cos( \frac{ \pi}{ 3 })+isin(\frac{ \pi }{ 3}))\]

Answer 4

nope, that's the polar form

Answer 5

Is this correct when the argument is between o and 2\[\pi \]?

Answer 6

there is mistake in ur polar form i think

Answer 7

which part is the mistake?

Answer 8

the angle

Answer 9

\(\alpha\)

Answer 10

\[\tan \prime(\sqrt{3})\]

Answer 11

\(tan^{-1}(-\sqrt{3})\)

Answer 12

Oh it was negative...

Answer 13

okay thank you!

Answer 14

so its
\(\Large 2[cos(300^{\circ})+isin(300^{\circ})]\)

Answer 15

\(\Large 2[cos(\dfrac{5\pi}{3})+isin(\dfrac{5\pi}{3})]\)