# principal vaue of a complex number

• December 18th 2010, 02:05 AM
killykilly
principal vaue of a complex number
I strugged for half an hour trying to find materials for this problem.

$(3+j4)^{1+2j}$

could you lead me to a good study material for this topic ?

thanks
• December 18th 2010, 02:35 AM
Prove It
Is this $\displaystyle (3+4j)^{1+2j}$?

If so, you should note that complex exponentiation is best done using the exponential form of the complex number.

$\displaystyle |3+4j| = \sqrt{3^2 + 4^2} = 5$

$\displaystyle \arg{(3+4j)} = \arctan{\frac{4}{3}}$.

So $\displaystyle 3 + 4j = 5e^{j\arctan{\frac{4}{3}}}$

$\displaystyle (3 + 4j)^{1+2j} = (5e^{j\arctan{\frac{4}{3}}})^{1+2j}$

$\displaystyle = 5^{1+2j}\,e^{j\arctan{\frac{4}{3}}(1 + 2j)}$

$\displaystyle = e^{\log{(5^{1+2j})}}\,e^{j\arctan{\frac{4}{3}}(1+2 j)}$

$\displaystyle = e^{(1+2j)\log{5}}\,e^{j\arctan{\frac{4}{3}}(1+2j)}$

$\displaystyle = e^{(1+2j)\log{5}+j\arctan{\frac{4}{3}}(1+2j)}$

$\displaystyle = e^{\log{5} + 2\log{5}j + j\arctan{\frac{4}{3} - 2\arctan{\frac{4}{3}}}$

$\displaystyle = e^{\log{5} - 2\arctan{\frac{4}{3}} + \left(2\log{5} + \arctan{\frac{4}{3}}\right)j}$

$\displaystyle = e^{\log{5}-2\arctan{\frac{4}{3}}}\,e^{\left(2\log{5} + \arctan{\frac{4}{3}}\right)j}$

$\displaystyle = e^{\log{5}- 2\arctan{\frac{4}{3}}}\left[\cos{\left(2\log{5} + \arctan{\frac{4}{3}}\right)} + i\sin{\left(2\log{5} + \arctan{\frac{4}{3}} \right)}\right]$

$\displaystyle = e^{\log{5} - 2\arctan{\frac{4}{3}}}\cos{\left(2\log{5} + \arctan{\frac{4}{3}}\right)} + i\,e^{\log{5} - 2\arctan{\frac{4}{3}}}\sin{\left(2\log{5} + \arctan{\frac{4}{3}}\right)}$

$\displaystyle = 5e^{-2\arctan{\frac{4}{3}}}\cos{\left(2\log{5} + \arctan{\frac{4}{3}}\right)} + 5i\,e^{-2\arctan{\frac{4}{3}}}\sin{\left(2\log{5} + \arctan{\frac{4}{3}}\right)}$

PHEW!!!
• December 18th 2010, 03:45 AM
killykilly
wow. never thought that it's that complicated. thank you !

the source where i got it from says that the answer is -0.42 - j0.66

also, my python calculator also says

>>> pow((3+4j),(1+2j))
(-0.41981317556195746-0.6604516942073323j)

how can we prove it ?
• December 18th 2010, 03:47 AM
Prove It
Why don't you try evaluating the real and imaginary parts of the answer I've posted?