# Thread: Write in a + bi Form

1. ## Write in a + bi Form

Write the expression in the standard form a + bi.

(1 - sqrt{5} i)^8

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

(1 - sqrt{5} i)^8
Let $\displaystyle z = 1 - \sqrt{5} \, i$.

You need to find $\displaystyle z^8$.

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

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

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

3. ## great....

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

You need to find $\displaystyle z^8$.

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

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

Therefore $\displaystyle (z^4)^2 = z^8 = \, ....$
A great start....

4. Originally Posted by magentarita
Write the expression in the standard form a + bi. (1 - sqrt{5} i)^8
$\displaystyle \alpha = \arctan \left( { - \sqrt 5 } \right)$
$\displaystyle z = \sqrt 6 e^{i\alpha }$
$\displaystyle z^8 = \left( {\sqrt 6 } \right)^8 e^{i8\alpha }$

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

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

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?

6. ## yes....

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.