Write ##5-3i## in the polar form ##re^\left(i\theta\right)##

[Mutatis]

Homework Statement

Write $5-3i$ in the polar form $re^\left(i\theta\right)$.

Homework Equations

$$
|z|=\sqrt {a^2+b^2}
$$

The Attempt at a Solution

First I've found the absolute value of $z$:
$$ |z|=\sqrt {5^2+3^2}=\sqrt {34} $$.
Next, I've found $$ \sin(\theta) = \frac {-3} {\sqrt {34}} \\ \cos(\theta) = \frac {5} {\sqrt {34}}.$$ I don't know what to do next because I haven't found any exact value from these numbers...

 

[SammyS]

Homework Statement

Write $5-3i$ in the polar form $re^\left(i\theta\right)$.

Homework Equations

$$
|z|=\sqrt {a^2+b^2}
$$

The Attempt at a Solution

First I've found the absolute value of $z$:
$$ |z|=\sqrt {5^2+3^2}=\sqrt {34} $$.Next, I've found $$ \sin(\theta) = \frac {-3} {\sqrt {34}} \\ \cos(\theta) = \frac {5} {\sqrt {34}}.$$ I don't know what to do next because I haven't found any exact value from these numbers...

You may want to use $\tan(\theta)$ to get a simpler expression. One without a radical.

Solve for $\theta$.

Take care to get $\theta$ into the correct quadrant.

 

[Ray Vickson]

Homework Statement

Write $5-3i$ in the polar form $re^\left(i\theta\right)$.

Homework Equations

$$
|z|=\sqrt {a^2+b^2}
$$

The Attempt at a Solution

First I've found the absolute value of $z$:
$$ |z|=\sqrt {5^2+3^2}=\sqrt {34} $$.
Next, I've found $$ \sin(\theta) = \frac {-3} {\sqrt {34}} \\ \cos(\theta) = \frac {5} {\sqrt {34}}.$$ I don't know what to do next because I haven't found any exact value from these numbers...

You should not be bothered by not being able to give an exact "algebraic" value for $\theta.$ Most values of $\arcsin(u),$ $\arccos(v)$ and $\arctan(w)$ do not have nice formulas, even if you have nice formulas for $u,v,w.$ That is why numerical methods are often necessary. You may be able to give a numerical value of $\theta$ to 100 decimal places but never find an "algebraic" formula for it.

 

[Mutatis]

I did it the way you told me to. I'd wrote in terms of tg: $$ \tan \left( \frac {-3} 5 \right) \\ \theta = \operatorname {arctg} \left( \frac {-3} 5 \right) \\ z =\sqrt {34} \left[ \cos \left( \operatorname{arctg} \left( \frac {-3} 5 \right) \right) + i\sin \left( \operatorname {arctg} \left( \frac {-3} 5 \right) \right) \right] .$$ If this is right, thank you guys!

 

[RPinPA]

That's OK for this problem, but one caveat: For any value, there are two different angles 180 degrees ($\pi$ radians) apart with the same tangent. The arctan function will give you the one that is between $-\pi/2$ and $\pi/2$. Since $5 - 3i$ is in quadrant IV (between $-\pi/2$ and 0), that gives the right value here.

But for quadrants II and III, you'd have to add or subtract $\pi$, that is change the sign of the cosine and sine. For instance if your value was $-5 + 3i$, that would have the same tangent of (-3/5) but arctan(-3/5) is 180 degrees off.

 

[Ray Vickson]

I did it the way you told me to. I'd wrote in terms of tg: $$ \tan \left( \frac {-3} 5 \right) \\ \theta = \operatorname {arctg} \left( \frac {-3} 5 \right) \\ z =\sqrt {34} \left[ \cos \left( \operatorname{arctg} \left( \frac {-3} 5 \right) \right) + i\sin \left( \operatorname {arctg} \left( \frac {-3} 5 \right) \right) \right] .$$ If this is right, thank you guys!

All this last bit is unnecessary. If you regard the polar form as $z = r (\cos \theta + i \sin \theta)$ (as most sources do), then you might as well just type $5/\sqrt{34}$ and $-3/\sqrt{34}$ instead of your $\cos \arctan (-3/5)$ and $\sin \arctan (-3/5)$; in other words,
$$z = \sqrt{34} \left( \frac{5}{\sqrt{34}} - i \frac{3}{\sqrt{34}} \right)$$ However, if you regard the polar form as $z = r e^{i \theta}$ (as some people do, and your original question asked) then you need to find $\theta.$ In fact, $\theta = \arctan(-3/5) = ?$ some (approximate) numerical value. Substitute that numerical value into $r e^{i \theta}$ and you are done.

BTW: there is no need to write "\operatorname{arctg}..."in LaTeX; writing "\arctan ... " works perfectly well. It seems that LaTeX knows about inverse trig functions, but not about inverse hyperbolic functions such as "arcsinh", etc. In that case you need "\text{arcsinh} ... ", or "\operatorname{arcsinh} ..." if you prefer. (In fact, "operatorname" works better: it produces $\operatorname{arcsinh} x$ instead of the uglier $\text{arcsinh} x$ that "\text{arcsinh} x" produces (unless you deliberately insert a spacer).

 

[SammyS]

I did it the way you told me to. I'd wrote in terms of tg: $$ \tan \left( \frac {-3} 5 \right) $$

That should be $\ \tan(\theta) = \dfrac{-3}{5} ~.$

$$ \theta = \operatorname {arctg} \left( \frac {-3} 5 \right) \\ z =\sqrt {34} \left[ \cos \left( \operatorname{arctg} \left( \frac {-3} 5 \right) \right) + i\sin \left( \operatorname {arctg} \left( \frac {-3} 5 \right) \right) \right] .$$ If this is right, thank you guys!

I thought you wanted to put $\ 5-3i \$ in the polar form: $\ r e^{i\theta}$.
You have found $r$ and $\theta$. Plug them in.

So what if $\theta$ is $\arctan(-3/5)$.

 

[scottdave]

One other important thing to note - if you do convert the angle to a decimal number, make sure you are in radians mode. When you do $e^{i\theta}$, θ needs to be radians.

 

[Master1022]

Homework Statement

Write $5-3i$ in the polar form $re^\left(i\theta\right)$.

Hi. The other posters have already shown you how to get to the correct solution. I would just like to suggest that it is always worth drawing a diagram. Especially with complex numbers, it helps to see which quadrant you are in when calculating your arg(z) (as other posters have mentioned). Furthermore, it helps you to keep your working clear.

Hope this little tip helps.

 

[Mutatis]

Thank you very much!