# Write in a + bi Form

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• Oct 3rd 2008, 04:53 AM
magentarita
Write in a + bi Form
Write the expression in the standard form a + bi.

(1 - sqrt{5} i)^8
• Oct 3rd 2008, 05:06 AM
mr fantastic
Quote:

Originally Posted by magentarita
Write the expression in the standard form a + bi.

(1 - sqrt{5} i)^8

Let $z = 1 - \sqrt{5} \, i$.

You need to find $z^8$.

Note that $z^2 = (1 - \sqrt{5} \, i) (1 - \sqrt{5} \, i) = -4 - 2 \sqrt{5} \, i = -2 (2 + \sqrt{5} \, i)$.

Therefore $(z^2)^2 = z^4 = [ -2 (2 + \sqrt{5} \, i) ]^2 = 4 (2 + \sqrt{5} \, i)^2 = \, ....$

Therefore $(z^4)^2 = z^8 = \, ....$
• Oct 3rd 2008, 04:51 PM
magentarita
great....
Quote:

Originally Posted by mr fantastic
Let $z = 1 - \sqrt{5} \, i$.

You need to find $z^8$.

Note that $z^2 = (1 - \sqrt{5} \, i) (1 - \sqrt{5} \, i) = -4 - 2 \sqrt{5} \, i = -2 (2 + \sqrt{5} \, i)$.

Therefore $(z^2)^2 = z^4 = [ -2 (2 + \sqrt{5} \, i) ]^2 = 4 (2 + \sqrt{5} \, i)^2 = \, ....$

Therefore $(z^4)^2 = z^8 = \, ....$

A great start....
• Oct 3rd 2008, 06:41 PM
Plato
Quote:

Originally Posted by magentarita
Write the expression in the standard form a + bi. (1 - sqrt{5} i)^8

$\alpha = \arctan \left( { - \sqrt 5 } \right)$
$z = \sqrt 6 e^{i\alpha }$
$z^8 = \left( {\sqrt 6 } \right)^8 e^{i8\alpha }$

YOU should know how to continue!
If not then you have no reason to ask this question!
• Oct 3rd 2008, 06:55 PM
Jhevon
Quote:

Originally Posted by Plato
$\alpha = \arctan \left( { - \sqrt 5 } \right)$
$z = \sqrt 6 e^{i\alpha }$
$z^8 = \left( {\sqrt 6 } \right)^8 e^{i8\alpha }$

YOU should know how to continue!
If not then you have no reason to ask this question!

are students usually taught how to write and manipulate the polar form of complex numbers in precalculus?
• Oct 4th 2008, 04:02 AM
magentarita
yes....
Quote:

Originally Posted by Jhevon
are students usually taught how to write and manipulate the polar form of complex numbers in precalculus?

It's a precalculus course in preparation for calculus 1 and an upcoming state exam.