complex number

**thereddevils** · May 20th 2009, 12:52 AM

Given that z=-2+i , express $z+\frac{1}{\sqrt{z}}$ in the form of a+bi

**mathsquest** · May 20th 2009, 01:26 AM

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.

**thereddevils** · May 20th 2009, 03:41 AM

Is there a simpler way of doing this ?

This is what i tried :

$\sqrt{-2+i}=a+bi$

$-2+i=a^2-b^2+2abi$

$-2=a^2-b^2$ and $2ab=1$

i am not sure whether i am correct up to this step but after solving , i get weird values for a and b ..

**mathsquest** · May 21st 2009, 08:53 PM

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!

**stapel** · May 22nd 2009, 07:57 AM

Originally Posted by thereddevils

Given that z=-2+i , express $z+\frac{1}{\sqrt{z}}$ in the form of a+bi

Just do the simplification:

$-2\, +\, i\, +\, \frac{1}{-2\, +\, i}$

$-2\, +\, i\, +\, \left(\frac{1}{-2\, +\, i}\right)\left(\frac{-2\, -\, i}{-2\, -\, i}\right)$

$-2\, +\, i\, +\, \frac{-2\, -\, i}{(-2)^2\, -\, i^2}$

$-2\, +\, i\, +\, \frac{-2\, -\, i}{4\, +\, 1}$

$-2\, +\, i\, +\, \frac{-2\, -\, i}{5}$

$-2\, -\, \frac{2}{5}\, +\, i\, -\, \frac{1}{5}i$

$-\frac{12}{5}\, +\, \frac{4}{5}i$

**mathsquest** · May 22nd 2009, 08:13 PM

You seem to have solved for Z + 1/Z and not Z + 1/sqrt Z

**Math's-only-a-game** · May 25th 2009, 11:13 AM

Originally Posted by thereddevils

Given that z=-2+i , express $z+\frac{1}{\sqrt{z}}$ in the form of a+bi

$ \sqrt z = \sqrt { - 2 + i} = a + bi, $

$ \therefore - 2 + i = (a + bi)^2 = a^2 + 2abi - b^2 ........\langle 1\rangle $
$ a^2 - b^2 = - 2,2ab = 1 \Rightarrow b^2 = \frac{1} {{4a^2 }} $

$ \text{Substitute into }\langle 1\rangle \text{: }4a^4 + 8a^2 - 1 = 4(a^2 )^2 + 8a^2 - 1\text{ }.........\langle \text{2}\rangle $

$ \text{Put }a^2 = x \Rightarrow \langle \text{2}\rangle = 4x^2 + 8x - 1 $
$ \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)} $
$ \text{From }2ab = 1\text{, }b = \frac{1} {{2a}} = \frac{1} {2}\left( { - 1 + \frac{1} {2}\sqrt 5 } \right)^{ - \tfrac{1} {2}} $

$ \therefore z = - 2 + i = (a + bi)^2 = $

$ = \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 $

$ \text{and }z + \frac{1} {{\sqrt z }} = $
$ = \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 + $
$ + \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} $

$ \text{which we need to reduce in the form of }a + bi.\text{ Any hints are highly appreciated}\text{.}$

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