Complex number real to polar form

[ronald]

I have this equation z=3-4i
so i find r=squareroot(x^2+y^2)
r=squareroot(3^2+4^2)
r=squareroot(25)
r= 5

then in its polar form

theta=tan-1(-4/3)=180degree-tan-1(4/3)=126.87=2.2143 radians
so its form are z=5e^i2.2143

question how if z^3=3-4i...how to solve if z^3.

Thanks.

 

[pka]

I have this equation z=3-4i
so i find r=squareroot(x^2+y^2)
r=squareroot(3^2+4^2)
r=squareroot(25)
r= 5.

You did post this in the advanced forum. So here is the answer at that level.
Use \(\displaystyle \exp \left( {i\theta } \right) = \cos (\theta ) + i\sin (\theta ) \).

For this problem the argument is \(\displaystyle \theta = \arctan \left( {\frac{{ - 4}}{3}} \right)\).
Thus one root is \(\displaystyle \sqrt[3]{5}\exp \left( {\frac{\theta }{3}} \right)\)

The other two roots are \(\displaystyle \sqrt[3]{5}\exp \left( {\frac{\theta }{3}} \right) \cdot \exp \left( {\frac{{2\pi k}}{3}} \right),\;k = 1,2 \)

 

[ronald]

You did post this in the advanced forum. So here is the answer at that level.
Use \(\displaystyle \exp \left( {i\theta } \right) = \cos (\theta ) + i\sin (\theta ) \).

For this problem the argument is \(\displaystyle \theta = \arctan \left( {\frac{{ - 4}}{3}} \right)\).
Thus one root is \(\displaystyle \sqrt[3]{5}\exp \left( {\frac{\theta }{3}} \right)\)

The other two roots are \(\displaystyle \sqrt[3]{5}\exp \left( {\frac{\theta }{3}} \right) \cdot \exp \left( {\frac{{2\pi k}}{3}} \right),\;k = 1,2 \)

is a is the sqrt of z. thanks for this formulae.

 

[pka]

is a is the sqrt of z. thanks for this formulae.

\(\displaystyle \sqrt[3]{5}\) is the cube root of 5 not the square root.