# Math Help - Finding roots in a + ib form

1. ## Finding roots in a + ib form

Express the following in the form
a + ib. Give all values and make a polar plot of
the points or the vector vectors that represent your result.
(a)
(9i)^(1/2)

is this the answer? =3*(cos pi/2 + i sin pi/2)
=3i

so how do we represent this on polar plot?

(b)(-64i)^(1/4)=(64^0.25)*(cos theta/4 + i sin theta/4)

here im getting theta= -pi and +pi so which one do we take?

2. Originally Posted by sandy
Express the following in the form
a + ib. Give all values and make a polar plot of
the points or the vector vectors that represent your result.
(a)
(9i)^(1/2)

is this the answer? =3*(cos pi/2 + i sin pi/2)
=3i
obviously not. $(9i)^{1/2} \ne 3i$, since if you square $3i$ you get $-9$, not $9i$

Let $(9i)^{1/2} = a + ib$

$\Rightarrow 9i = (a + ib)^2 = a^2 - b^2 + 2abi$

Thus, $a^2 - b^2 = 0$ ........(1) and

$2ab = 9$ ...............(2)

Now solve this system simultaneously, and note that $a,b \in \mathbb R$

3. Originally Posted by Jhevon
obviously not. $(9i)^{1/2} \ne 3i$, since if you square $3i$ you get $-9$, not $9i$

Let $(9i)^{1/2} = a + ib$

$\Rightarrow 9i = (a + ib)^2 = a^2 - b^2 + 2abi$

Thus, $a^2 - b^2 = 0$ ........(1) and

$2ab = 9$ ...............(2)

Now solve this system simultaneously, and note that $a,b \in \mathbb R$
Isn't this a multi-branch function?

4. Originally Posted by Drexel28
Isn't this a multi-branch function?
i don't think they want to get that complicated here. it seems they are just asking for the 2 square roots.

5. Originally Posted by sandy
Express the following in the form
a + ib. Give all values and make a polar plot of
the points or the vector vectors that represent your result.
(a)
(9i)^(1/2)

is this the answer? =3*(cos pi/2 + i sin pi/2)
=3i

so how do we represent this on polar plot?

(b)(-64i)^(1/4)=(64^0.25)*(cos theta/4 + i sin theta/4)

here im getting theta= -pi and +pi so which one do we take?
You need to remember that there are exactly two square roots, and they are evenly spaced around the unit circle. In other words, they differ by an angle of $\pi$.

Convert $9i$ to polars:

$|9i| = 9$

$\arg{(9i)} = \frac{\pi}{2}$.

So $9i = 9\,\textrm{cis}\,\frac{\pi}{2}$.

Therefore $(9i)^{\frac{1}{2}} = \left(9\,\textrm{cis}\,\frac{\pi}{2}\right)^{\frac {1}{2}}$

$= 9^{\frac{1}{2}}\,\textrm{cis}\,\left(\frac{1}{2}\c dot \frac{\pi}{2}\right)$ By DeMoivre's Theorem

$= 3\,\textrm{cis}\,\frac{\pi}{4}$.

And remembering that the other square root only differs by an angle of $\pi$, that means the other square root is

$3 \,\textrm{cis}\,\frac{5\pi}{4}$.

Converting back to Cartesians:

$3\,\textrm{cis}\,\frac{\pi}{4} = 3\cos{\frac{\pi}{4}} + 3i\sin{\frac{\pi}{4}}$

$= \frac{3\sqrt{2}}{2} + i\,\frac{3\sqrt{2}}{2}$.

$3\,\textrm{cis}\,\frac{5\pi}{4} = 3\cos{\frac{5\pi}{4}} + 3i\sin{\frac{5\pi}{4}}$

$= -\frac{3\sqrt{2}}{2} - i\,\frac{3\sqrt{2}}{2}$.

6. Originally Posted by Prove It
You need to remember that there are exactly two square roots, and they are evenly spaced around the unit circle. In other words, they differ by an angle of $\pi$.

Convert $9i$ to polars:

$|9i| = 9$

$\arg{(9i)} = \frac{\pi}{2}$.

So $9i = 9\,\textrm{cis}\,\frac{\pi}{2}$.

Therefore $(9i)^{\frac{1}{2}} = \left(9\,\textrm{cis}\,\frac{\pi}{2}\right)^{\frac {1}{2}}$

$= 9^{\frac{1}{2}}\,\textrm{cis}\,\left(\frac{1}{2}\c dot \frac{\pi}{2}\right)$ By DeMoivre's Theorem

$= 3\,\textrm{cis}\,\frac{\pi}{4}$.

And remembering that the other square root only differs by an angle of $\pi$, that means the other square root is

$3 \,\textrm{cis}\,\frac{5\pi}{4}$.

Converting back to Cartesians:

$3\,\textrm{cis}\,\frac{\pi}{4} = 3\cos{\frac{\pi}{4}} + 3i\sin{\frac{\pi}{4}}$

$= \frac{3\sqrt{2}}{2} + i\,\frac{3\sqrt{2}}{2}$.

$3\,\textrm{cis}\,\frac{5\pi}{4} = 3\cos{\frac{5\pi}{4}} + 3i\sin{\frac{5\pi}{4}}$

$= -\frac{3\sqrt{2}}{2} - i\,\frac{3\sqrt{2}}{2}$.
thanks this seems to be reasonable compared to what i done

but what is the arg(-64i) ??

7. Originally Posted by sandy
what is the arg(-64i) ??
If $r>0$ then $\text{Arg}(-ri)=\frac{-\pi}{2}$.

8. Originally Posted by sandy
thanks this seems to be reasonable compared to what i done

but what is the arg(-64i) ??
Sorry, but this is absolutley basic. Did you try drawing an Argand diagram and using the definition of arg?

9. Originally Posted by mr fantastic
Sorry, but this is absolutley basic. Did you try drawing an Argand diagram and using the definition of arg?
i done the diagram its either -pi/2 or -pi/2 + 2kpi= 3pi/2

10. Originally Posted by sandy
i done the diagram its either -pi/2 or -pi/2 + 2kpi= 3pi/2

$\arg{z} = -\frac{\pi}{2} + 2\pi k, k \in \mathbf{Z}$

but the Principle Argument

$\textrm{Arg}\,{z} = -\frac{\pi}{2}$

since $-\pi < \textrm{Arg}\,{z} \leq \pi$.

11. Originally Posted by Prove It
$\arg{z} = -\frac{\pi}{2} + 2\pi k, k \in \mathbf{Z}$

but the Principle Argument

$\textrm{Arg}\,{z} = -\frac{\pi}{2}$

since $-\pi < \textrm{Arg}\,{z} \leq \pi$.
thanks alot so i did use - pi/2 so can you please go and check if the answer is right the one i posted please