# polar and cartesian coordinates

• Jan 7th 2009, 05:43 PM
dixie
polar and cartesian coordinates
express values of the following expressions in both polar and cartesian coordinates and plot them
a) square root of i
b) ((1+i)/square root of 2) to the 25 power

thanks, the i are confusing me
• Jan 7th 2009, 07:35 PM
Mush
Quote:

Originally Posted by dixie
express values of the following expressions in both polar and cartesian coordinates and plot them
a) square root of i
b) ((1+i)/square root of 2) to the 25 power

thanks, the i are confusing me

1) $a+bi = \sqrt{i}$

$(a+bi)^2 = i$

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

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

Compare imaginary parts $2abi = i$

$2ab = 1$

$ab = \frac{1}{2}$

$a = \frac{1}{2b}$

Now compare real parts

$( \frac{1}{2b})^2-b^2 = 0$

$\frac{1}{4b^2} = b^2$

$\frac{1}{4} = b^4$

$\pm (\frac{1}{4})^{\frac{1}{4}} = b$

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

$a = \frac{1}{2}.\frac{2}{\sqrt{2}}$

$a = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

$\therefore \sqrt{i} = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}$

b) $a+bi = (\frac{1+i}{\sqrt{2}})^{25}$

$(a+bi)^{\frac{1}{25}} = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i)$

$(a+bi)^{\frac{1}{25}} = (\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i)$

And the RHS, as we seen in question a), is $\sqrt{i}$

$(a+bi)^{\frac{1}{25}} =\sqrt{i}$

$a+bi =\sqrt{i^{25}}$

A quick check on the calculate says that $i^{25} = i$.

$a+bi =\sqrt{i} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i$

Both were identical. I'll leave you to put them in polar form ;)