# Complex Complex

• Mar 17th 2010, 11:50 PM
banku12
Complex Complex
if a is a complex number which satisfy $\displaystyle ia^3+a^2-a+1=0$

then find $\displaystyle \left | a \right |$?

one way is to put a=x+iy

any other short way ?
• Mar 18th 2010, 08:10 AM
Soroban
Hello, banku12!

Is there a typo?
If the last term is $\displaystyle i$, I can solve it.

Quote:

If a is a complex number which satisfies: .$\displaystyle ia^3+a^2-a + {\color{red}i}\:=\:0$

then find: .$\displaystyle |a|$

Divide by $\displaystyle i\!:\;\;a^3 + \frac{a^2}{i} - \frac{a}{i} + \frac{i}{i} \:=\:0 \quad\Rightarrow\quad a^3 - ia^2 + ia + 1 \:=\:0$

$\displaystyle \begin{array}{ccccc}\text{We have:} & a^3 - ia^2 + ia - i^2 &=& 0 \\ \\ \text{Factor:} & a^2(a-i) + i(a-i) &=& 0 \\ \\ \text{Factor:} & (a-i)(a^2+i) &=& 0 \end{array}$

. . Hence: .$\displaystyle a \;=\;i,\;\pm\sqrt{i}$

Therefore, for all roots: .$\displaystyle |a|\:=\:1$

• Mar 18th 2010, 09:17 AM
banku12
No..actually there is no typo..

btw is it not possible to compute mod a if last term in the given exp is 1 ?