# Finding roots in a + ib form

#### 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?

#### Jhevon

MHF Hall of Honor
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. $$\displaystyle (9i)^{1/2} \ne 3i$$, since if you square $$\displaystyle 3i$$ you get $$\displaystyle -9$$, not $$\displaystyle 9i$$

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

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

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

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

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

#### Drexel28

MHF Hall of Honor
obviously not. $$\displaystyle (9i)^{1/2} \ne 3i$$, since if you square $$\displaystyle 3i$$ you get $$\displaystyle -9$$, not $$\displaystyle 9i$$

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

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

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

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

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

#### Jhevon

MHF Hall of Honor
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.

#### Prove It

MHF Helper
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 $$\displaystyle \pi$$.

Convert $$\displaystyle 9i$$ to polars:

$$\displaystyle |9i| = 9$$

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

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

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

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

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

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

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

Converting back to Cartesians:

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

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

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

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

#### sandy

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 $$\displaystyle \pi$$.

Convert $$\displaystyle 9i$$ to polars:

$$\displaystyle |9i| = 9$$

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

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

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

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

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

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

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

Converting back to Cartesians:

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

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

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

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

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

#### Plato

MHF Helper
what is the arg(-64i) ??
If $$\displaystyle r>0$$ then $$\displaystyle \text{Arg}(-ri)=\frac{-\pi}{2}$$.

#### mr fantastic

MHF Hall of Fame
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?

#### sandy

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

#### Prove It

MHF Helper
i done the diagram its either -pi/2 or -pi/2 + 2kpi= 3pi/2

$$\displaystyle \arg{z} = -\frac{\pi}{2} + 2\pi k, k \in \mathbf{Z}$$
$$\displaystyle \textrm{Arg}\,{z} = -\frac{\pi}{2}$$
since $$\displaystyle -\pi < \textrm{Arg}\,{z} \leq \pi$$.