# For exam tomorrow. Complex numbers.

• Jun 1st 2008, 02:26 PM
AshleyT
For exam tomorrow. Complex numbers.
Arg, this mark scheme is...rubbish...

Okey, use the fact the suare root of 16 + 30i is (-5 - 3i) and (5 + 3i)

To work out $z^2 - 2z - (15 + 30i)$

Using the formula i got to this stage and got stuck =(

$2 + or - \frac{\sqrt{64 + 120i}}{2}$

Now i notice if you \ the fraction by 4 you get 16 + 30i which your meant to use...but you can't just do that so uh..wt..

Any help, i'd appreciate loads! Thanks (answer is 6 + 3i and -4 - 3i)

tytyty
• Jun 1st 2008, 02:35 PM
Moo
Hi !

Quote:

Originally Posted by AshleyT
Arg, this mark scheme is...rubbish...

Okey, use the fact the suare root of 16 + 30i is (-5 - 3i) and (5 + 3i)

To work out $z^2 - 2z - (15 + 30i)$

Using the formula i got to this stage and got stuck =(

$2 + or - \frac{\sqrt{64 + 120i}}{2}$

Now i notice if you \ the fraction by 4 you get 16 + 30i which your meant to use...but you can't just do that so uh..wt..

Any help, i'd appreciate loads! Thanks (answer is 6 + 3i and -4 - 3i)

tytyty

It's nice that you noticed that (Wink)

$\sqrt{64+120i}=\sqrt{4(16+30i)}=2 \sqrt{16+30i}$

I think that's all you need here :D

Just show your work if you have further difficulties, it'll be rather better to help you understand :)
• Jun 1st 2008, 02:45 PM
AshleyT
Quote:

Originally Posted by Moo
Hi !

It's nice that you noticed that (Wink)

$\sqrt{64+120i}=\sqrt{4(16+30i)}=2 \sqrt{16+30i}$

I think that's all you need here :D

Just show your work if you have further difficulties, it'll be rather better to help you understand :)

O yea! Can't believe missed that...

Thank-you very much!!