representing this complex fracture into exponent

• Jan 20th 2010, 02:39 AM
transgalactic
representing this complex fracture into exponent
$\frac{2i}{2+i}$

i don know how to separate the complex and imaginary part of this fracture?
• Jan 20th 2010, 02:59 AM
tonio
Quote:

Originally Posted by transgalactic
$\frac{2i}{2+i}$

i don know how to separate the complex and imaginary part of this fracture?

Twice you wrote "fracture": perhaps you meant "fraction"? Anyway, multiplying by the denominator's conjugate:

$\frac{1}{2+i}=\frac{2-i}{5}= \frac{2}{5}-\frac{1}{5}\,i$ , and now just multiply this by $2i$ ...

Tonio
• Jan 20th 2010, 04:38 AM
Prove It
Quote:

Originally Posted by transgalactic
$\frac{2i}{2+i}$

i don know how to separate the complex and imaginary part of this fracture?

$\frac{2i}{2 + i} = \frac{2i(2 - i)}{(2 + i)(2 - i)}$

$= \frac{4i - 2i^2}{4 - i^2}$

$= \frac{2 + 4i}{5}$

$= \frac{2}{5} + \frac{4}{5}i$.

Now putting into polar form...

$|z| = \sqrt{\left(\frac{2}{5}\right)^2 + \left(\frac{4}{5}\right)^2}$

$= \sqrt{\frac{4}{25} + \frac{16}{25}}$

$= \sqrt{\frac{20}{25}}$

$= \frac{2\sqrt{5}}{5}$.

$\theta = \arctan{\frac{\frac{4}{5}}{\frac{2}{5}}}$

$= \arctan{2}$.

So $z = \frac{2\sqrt{5}}{5}\,\textrm{cis}\,\arctan{2}$

$= \frac{2\sqrt{5}}{5}e^{i\arctan{2}}$.
• Jan 20th 2010, 05:05 AM
transgalactic
thanks:)