Imaginary Numbers?

• Mar 21st 2010, 05:35 PM
matherrsingerr
Imaginary Numbers?
I don't understand these math problems and i have get them done by tomorrow. btw they are imaginary numbers, so i squared is replaced with 1, i believe.

(7+7i)^2

2i/1+2i

-6+4i/1+4i

• Mar 21st 2010, 05:43 PM
Plato
Quote:

Originally Posted by matherrsingerr
(7+7i)^2

2i/1+2i
-6+4i/1+4i

For the first one:$\displaystyle (a+bi)^2=(a^2-b^2)+(2ab)i$

For the second just understand that $\displaystyle \frac{1}{a+bi}=\frac{a-bi}{a^2+b^2}$.
• Mar 22nd 2010, 03:12 AM
mr fantastic
Quote:

Originally Posted by matherrsingerr
I don't understand these math problems and i have get them done by tomorrow. btw they are imaginary numbers, so i squared is replaced with 1, i believe.

[snip]

No. i^2 = -1.
• Mar 22nd 2010, 08:51 AM
Gojinn
$\displaystyle (7+7i)^2$ - Treat it like a normal polynomial, but whn you get an $\displaystyle i^2$, replace with $\displaystyle -1$.

$\displaystyle \frac{2i}{1+2i}$ - Use the complex conjugate of the denominator and multiply both top and bottom by that conjugate. (Complex Conjugate is given by taking the complex statement, keeping the real parts idenitcal and changing the sign of the negative part).

Example: $\displaystyle \frac{1+4i}{2+3i}$ -> $\displaystyle \frac{1+4i}{2+3i} . \frac{2-3i}{2-3i}$ -> $\displaystyle \frac{(1+4i)(2-3i)}{(2+3i)(2-3i)}$ -> $\displaystyle \frac{14+5i}{7}$ -> $\displaystyle 2 + \frac{5}{7}i$