Author Topic: Need help solving equations with the square root of z (Read 804 times) Tweet Share

Ax3 · « **on:** March 06, 2010, 01:54:38 pm »

Hey guys i'm looking over the worked solutions to an example in the text (Find $\sqrt{3+4i}$ in cartesian form).

I can follow right along with the problem, where $\mathbf{x}=\pm2$ , and $\mathbf{y}=\pm1$ , but then the answer skips right to
"Therefore, the two roots of $z_1, z_2$ are $z_1=2+\mathbf{i}$ and $z_2=-2-\mathbf{i}$
"So $\sqrt{3+4i}$ is $2+\mathbf{i}$ or $z_2=-2-\mathbf{i}$ .

Where does the $i$ get added in?!

Martoman · « **Reply #1 on:** March 06, 2010, 03:46:09 pm »

Ah ok, this requires the observation that any complex number $z = x+iy$ .

Then if x = 2, y =1, so $z = 2 + i\times 1 = 2+ i$ and x = -2, y = -1 $z = -2 - i \times -1 = -2 -i$

Ax3 · « **Reply #2 on:** March 06, 2010, 03:47:40 pm »

Oh right right! Thanks man i appreciate it!

TrueTears · « **Reply #3 on:** March 07, 2010, 04:09:26 pm »

Quote from: Ax3 on March 06, 2010, 01:54:38 pm

Hey guys i'm looking over the worked solutions to an example in the text (Find $\sqrt{3+4i}$ in cartesian form).

I can follow right along with the problem, where $\mathbf{x}=\pm2$ , and $\mathbf{y}=\pm1$ , but then the answer skips right to
"Therefore, the two roots of $z_1, z_2$ are $z_1=2+\mathbf{i}$ and $z_2=-2-\mathbf{i}$
"So $\sqrt{3+4i}$ is $2+\mathbf{i}$ or $z_2=-2-\mathbf{i}$ .

Where does the $i$ get added in?!

You can also do let $z = \sqrt{3+4i}$

z^2 = 3+4i and then solve it using de moivres theorem (yeah I know a tan^-1[4/3] comes up but just another method xD)

Search the forums now!

We have moved!

Author Topic: Need help solving equations with the square root of z (Read 804 times) Tweet Share

Ax3

Need help solving equations with the square root of z

Martoman

Re: Need help solving equations with the square root of z

Ax3

Re: Need help solving equations with the square root of z

TrueTears

Re: Need help solving equations with the square root of z

Recent Posts

User:
Password: