trigonometric form of complex number

[fallenstar223]

i have a test on this tomorrow. and i have a quick question.

the problem gives a graph with a real and imaginary axis
and says that z=5-2i

i know that you plug it into r[cos0 + isin 0]
my teacher shortens it to rcis0

i have r = sqrt 29
but im having trouble finding what theta (0) is.
i know that 0 is suppose to be the 2nd function tangent of b/a
but when i do that for this problem im getting 2nd function tan of (-2/5) which i get as -22 degrees
which is not the same answer as in the back of the book.

any help would be greatly appreciated[/code]

 

[pka]

The real problem in converting to polar form is finding the argument (the angle).
Here is a sure fire way:
\(\displaystyle \L
Arg(z = x + iy) = \left\{ {\begin{array}{lc}
{\arctan \left( {\frac{y}{x}} \right),} & {x > 0} \\
{\pi + \arctan \left( {\frac{y}{x}} \right),} & {x < 0\;\& \;y > 0} \\
{ - \pi + \arctan \left( {\frac{y}{x}} \right),} & {x < 0\;\& \;y < 0} \\
\end{array}} \right.\)

Many calculators and computer algebra systems have built in functions to do this. The above can be programmed into a calculator.

 

[soroban]

Hello, fallenstar223!

I think it's a simple error . . .

The problem gives a graph with a real and imaginary axis
and says that: \(\displaystyle \,z\:=\:5\,-\,2i\)

i know that you plug it into: \(\displaystyle \:r(\cos\theta\,+\,i\cdot\sin\theta)\)

i have: \(\displaystyle \,r\,=\,\sqrt{29}\)

but im having trouble finding what \(\displaystyle \theta\) is.

i know that \(\displaystyle \theta\) is suppose to be \(\displaystyle \tan^{-1}\left(\frac{b}{a}\right)\)

but when i do that for this problem i'm getting: \(\displaystyle \,\tan^{-1}\left(-\frac{2}{5}\right) \:=\:-22^o\)

which is not the same answer as in the back of the book. . I'm not surprised . . .

At this level of mathematics, angles are usually given in radians.

Switch your calculator to radian mode.

Or convert your answer: \(\displaystyle \:-21.80140949^o\,\times\,\frac{\pi}{180^o}\;=\;-0.380506377\text{ radians.}\)