Complex numbers, cartesian to polar

[Anthonyk2013]

Wondering if I m right or wrong?

I have to convert 4-j² and -3+j⁵ to polar form

4-(-1)=4+1

r=square root of 4²+1²=square root of 17=4.123

A=Tan^-1(1/4)=14.03⁰

4.123angle 14.03⁰



-3+j⁵=-3+j

r=square root of -3²+1²=square root of 10=3.16

A=Tan^-1(1/-3)=-18.43⁰

Angle=360⁰-(-18.43)=378.43⁰

3.16angle 373.43⁰

SORRY CANT FIND THE SQUARE ROOT OR ANGLE SYMBOL

 

[pka]

Wondering if I m right or wrong?
I have to convert 4-j² -3+j⁵ to polar form

If the question is \(\displaystyle 4-i^2-3+i^5\) then you are incorrect.

For \(\displaystyle i^2=-1\) so \(\displaystyle -i^2=1\). Thus \(\displaystyle 4-i^2-3+i^5=2+i\).

 

[Anthonyk2013]

If the question is \(\displaystyle 4-i^2-3+i^5\) then you are incorrect.

For \(\displaystyle i^2=-1\) so \(\displaystyle -i^2=1\). Thus \(\displaystyle 4-i^2-3+i^5=2+i\).

No they are separate 4-j²and -3+j⁵

 

[pka]

No they are separate 4-j²and -3+j⁵

\(\displaystyle 4-i^2=5\) that is in polar form.

\(\displaystyle -3+i^5=-3+i=\sqrt {10} \left[ {\cos \left( {\pi - \arctan \left( {\frac{1}{3}} \right)} \right) + i\sin \left( {\pi - \arctan \left( {\frac{1}{3}} \right)} \right)} \right] \)

 

[Anthonyk2013]

\(\displaystyle 4-i^2=5\) that is in polar form.

\(\displaystyle -3+i^5=-3+i=\sqrt {10} \left[ {\cos \left( {\pi - \arctan \left( {\frac{1}{3}} \right)} \right) + i\sin \left( {\pi - \arctan \left( {\frac{1}{3}} \right)} \right)} \right] \)

So 5 is my answer for 4-j²

What about second part

is this ok
-3+j⁵=-3+j

r=square root of -3²+1²=square root of 10=3.16

A=Tan^-1(1/-3)=-18.43⁰

Angle=360⁰-(-18.43)=378.43⁰

3.16angle 373.43⁰

 

[Quaid]

r = square root of -3²+1²

Hi Anthony.

Be careful with notation.

-3² is -9 because we do exponents before multiplication.

(-3)² is what you're thinking, so don't forget the grouping symbols.


Also, some of your values are not rounded correctly, if it matters.

SORRY CANT FIND THE SQUARE ROOT OR ANGLE SYMBOL

If you're using Windows, characters like √ and ° can be pasted from the Character Map.

Otherwise, you may simply type expressions like

sqrt(16)

sqrt[(-3)^2 + 1^2]

tan(45 deg)

arctan(-1/3) = -18.44 deg


Cheers :cool:

 

[Anthonyk2013]

Hi Anthony.

Be careful with notation.

-3² is -9 because we do exponents before multiplication.

(-3)² is what you're thinking, so don't forget the grouping symbols.


Also, some of your values are not rounded correctly, if it matters.





If you're using Windows, characters like √ and ° can be pasted from the Character Map.

Otherwise, you may simply type expressions like

sqrt(16)

sqrt[(-3)^2 + 1^2]

tan(45 deg)

arctan(-1/3) = -18.44 deg


Cheers :cool:

Thanks well do.

 

[pka]

sqrt[(-3)^2 + 1^2]
tan(45 deg)
arctan(-1/3) = -18.44 deg

@Quaid, actually that is incorrect.

I gave the correct answer in post #4.
\(\displaystyle -3+i^5=-3+i=\sqrt {10} \left[ {\cos \left( {\pi - \arctan \left( {\frac{1}{3}} \right)} \right) + i\sin \left( {\pi - \arctan \left( {\frac{1}{3}} \right)} \right)} \right] \).

Hardly an working mathematician still uses degrees.
\(\displaystyle \left( {\pi - \arctan \left( {\frac{1}{3}} \right)} \right)\sim 161.6^o\), See here.

Note that \(\displaystyle -3+i\in II\) so \(\displaystyle \dfrac{\pi}{2}<\text{ARG}(-3+i)<\pi~.\)

 

[Anthonyk2013]

@Quaid, actually that is incorrect.

I gave the correct answer in post #4.
\(\displaystyle -3+i^5=-3+i=\sqrt {10} \left[ {\cos \left( {\pi - \arctan \left( {\frac{1}{3}} \right)} \right) + i\sin \left( {\pi - \arctan \left( {\frac{1}{3}} \right)} \right)} \right] \).

Hardly an working mathematician still uses degrees.
\(\displaystyle \left( {\pi - \arctan \left( {\frac{1}{3}} \right)} \right)\sim 161.6^o\), See here.

Note that \(\displaystyle -3+i\in II\) so \(\displaystyle \dfrac{\pi}{2}<\text{ARG}(-3+i)<\pi~.\)

Thanks for your help but i'am not familiar with that method. I will research it thanks again.

 

[Quaid]

arctan(-1/3) = -18.44 deg

actually that is incorrect.

Hi pka:

You may have misread my post. The OP had a concern about not knowing how to post a degree symbol; I suggested that typing "deg" is acceptable, and the line in red above is one example.

Cheers :cool: