Question

Write the polar form of the complex number 3-2i, using θ such that 0 ≤ θ ≤ 360º. Round numerical values to the nearest hundredth.

Answer 1

you need two numbers to find this and put it in the form
\[3-2i=r(\cos(\theta)+i\sin(\theta))\] namely \(r\) and \(\theta\)

Answer 2

\(r\) is straightforward, it is
\[r=\sqrt{a^2+b^2}\] in your case \(a=3,b=2\)

Answer 3

well actually \(b=-2\) but it makes no difference when you square it
what do you get for \(r\) ?

Answer 4

2.2?
i think i did it wrong though..

Answer 5

yeah i think so
you need \(r=\sqrt{3^2+2^2}\)

Answer 6

3.6

Answer 7

i guess, it is actually \(\sqrt{13}\)

Answer 8

if you round to the nearest 100th it is \(3.61\)

Answer 9

now for \(\theta\)
since \(3-2i\) is in quadrant 4, you can use
\[\theta=\tan^{-1}(\frac{b}{a})=\tan^{-1}(-\frac{2}{3})\]

Answer 10

and of course a calculator to find it

Answer 11

-33.69

Answer 12

maybe if you are working in degrees
usually these are just numbers (radians) but if you are working in degrees then you are right

Answer 13

you get a final answer of
\[3-2i=3.61(\cos(-33.69)+i\sin(-33.69))\]

Answer 14

thank you!