complex number question

• May 20th 2009, 11:26 PM
thereddevils
complex number question
Find the square root of the following complex number :

$6+8i$

$\sqrt{6+8i}=a+bi$

$6+8i=a^2-b^2+2abi$

$
a^2-b^2=6 ----1
$

$2ab=8 ---- 2$

From 2 , $b=\frac{4}{a}$

Substitute into 1 : $a^2-\frac{16}{a^2}=6$ which gives me
$a=\pm2\sqrt{2}$

then $b=\pm\frac{2}{\sqrt{2}}$

did i get the correct answers for a and b ? THanks for helping .
• May 20th 2009, 11:31 PM
Moo
Quote:

Originally Posted by thereddevils
Find the square root of the following complex number :

$6+8i$

$\sqrt{6+8i}=a+bi$

$6+8i=a^2-b^2+2abi$

$
a^2-b^2=6 ----1
$

$2ab=8 ---- 2$

From 2 , $b=\frac{4}{a}$

Substitute into 1 : $a^2-\frac{16}{a^2}=6$ which gives me
$a=\pm2\sqrt{2}$

then $b=\pm\frac{2}{\sqrt{2}}$

did i get the correct answers for a and b ? THanks for helping .

Yup, looks good.
• May 22nd 2009, 01:21 AM
Math's-only-a-game
Quote:

Originally Posted by thereddevils
Find the square root of the following complex number :

$6+8i$

$\sqrt{6+8i}=a+bi$

$6+8i=a^2-b^2+2abi$

$
a^2-b^2=6 ----1
$

$2ab=8 ---- 2$

From 2 , $b=\frac{4}{a}$

Substitute into 1 : $a^2-\frac{16}{a^2}=6$ which gives me
$a=\pm2\sqrt{2}$

then $b=\pm\frac{2}{\sqrt{2}}$

did i get the correct answers for a and b ? THanks for helping .

I'm trying to solve a similar problem. Could you please show me how did you get 'a'??
• May 22nd 2009, 01:40 AM
Moo
Quote:

Originally Posted by Math's-only-a-game
I'm trying to solve a similar problem. Could you please show me how did you get 'a'??

Multiply each side of the equation by $a^2$ :
$(a^2)^2-6a^2-16=0$

Let $x=a^2$, giving $x^2-6x-16=0$
and solve the quadratic. But be aware that $x=a^2\geq 0$. So you ought to keep only the positive root.