Given that z=-2+i , express$\displaystyle z+\frac{1}{\sqrt{z}}$ in the form of a+bi
Write -2 + i as x + i y ; r = (x^2 + y^2)^ (1/2)
Z = x + i y = r e^ i theta = r ( cos theta + i sin theta) from De Moivre;s
Theorem
where sin theta = y / (x^2 + y^2)^ (1/2) and
cos theta = x / (x^2 + y^2)^ (1/2)
Just substitute for Z + 1/Z and remember sin(-theta) = - sin theta.
Is there a simpler way of doing this ?
This is what i tried :
$\displaystyle \sqrt{-2+i}=a+bi$
$\displaystyle -2+i=a^2-b^2+2abi $
$\displaystyle -2=a^2-b^2$ and $\displaystyle 2ab=1 $
i am not sure whether i am correct up to this step but after solving , i get weird values for a and b ..
Actually the problem is because there are roots of a complex no involved I think and we have to involve the unit circle etc.
I tried it too . I got 4a^4 + 8a^2 - 1 = 0 . Put a^2 = x and then solve
4x^2 + 8 x - 1 = 0 to get x = -1 +or- 1/2 sqrt 5
Choose the +ve value and x = 0.118 or a = 0.3435 . Similarly try for b.
We have to ignore the value which is -ve else we end up with a being a complex no starting from the assumption it is not!
Just do the simplification:
$\displaystyle -2\, +\, i\, +\, \frac{1}{-2\, +\, i}$
$\displaystyle -2\, +\, i\, +\, \left(\frac{1}{-2\, +\, i}\right)\left(\frac{-2\, -\, i}{-2\, -\, i}\right)$
$\displaystyle -2\, +\, i\, +\, \frac{-2\, -\, i}{(-2)^2\, -\, i^2}$
$\displaystyle -2\, +\, i\, +\, \frac{-2\, -\, i}{4\, +\, 1}$
$\displaystyle -2\, +\, i\, +\, \frac{-2\, -\, i}{5}$
$\displaystyle -2\, -\, \frac{2}{5}\, +\, i\, -\, \frac{1}{5}i$
$\displaystyle -\frac{12}{5}\, +\, \frac{4}{5}i$
$\displaystyle
\sqrt z = \sqrt { - 2 + i} = a + bi,
$
$\displaystyle
\therefore - 2 + i = (a + bi)^2 = a^2 + 2abi - b^2 ........\langle 1\rangle
$
$\displaystyle
a^2 - b^2 = - 2,2ab = 1 \Rightarrow b^2 = \frac{1}
{{4a^2 }}
$
$\displaystyle
\text{Substitute into }\langle 1\rangle \text{: }4a^4 + 8a^2 - 1 = 4(a^2 )^2 + 8a^2 - 1\text{ }.........\langle \text{2}\rangle
$
$\displaystyle
\text{Put }a^2 = x \Rightarrow \langle \text{2}\rangle = 4x^2 + 8x - 1
$
$\displaystyle
\text{Set }\langle \text{2}\rangle = 0\text{ and solve: }a^2 = - 1 + \frac{1}
{2}\sqrt 5 \Rightarrow a = \sqrt { - 1 + \frac{1}
{2}\sqrt 5 } \text{ (only take + ve part)}
$
$\displaystyle
\text{From }2ab = 1\text{, }b = \frac{1}
{{2a}} = \frac{1}
{2}\left( { - 1 + \frac{1}
{2}\sqrt 5 } \right)^{ - \tfrac{1}
{2}}
$
$\displaystyle
\therefore z = - 2 + i = (a + bi)^2 =
$
$\displaystyle
= \left[ {\left( { - 1 + \frac{1}
{2}\sqrt 5 } \right)^{\tfrac{1}
{2}} + \frac{1}
{2}\left( { - 1 + \frac{1}
{2}\sqrt 5 } \right)^{ - \tfrac{1}
{2}} i} \right]^2
$
$\displaystyle
\text{and }z + \frac{1}
{{\sqrt z }} =
$
$\displaystyle
= \left[ {\left( { - 1 + \frac{1}
{2}\sqrt 5 } \right)^{\tfrac{1}
{2}} + \frac{1}
{2}\left( { - 1 + \frac{1}
{2}\sqrt 5 } \right)^{ - \tfrac{1}
{2}} i} \right]^2 +
$
$\displaystyle
+ \left[ {\left( { - 1 + \frac{1}
{2}\sqrt 5 } \right)^{\tfrac{1}
{2}} + \frac{1}
{2}\left( { - 1 + \frac{1}
{2}\sqrt 5 } \right)^{ - \tfrac{1}
{2}} i} \right]^{ - 1}
$
$\displaystyle
\text{which we need to reduce in the form of }a + bi.\text{ Any hints are highly appreciated}\text{.}$