z = re^i(theta)

• Nov 22nd 2007, 08:41 PM
ohhhh
z = re^i(theta)
I have noooooo idea how to do these types of problems. I missed the lecture when we were on this topic and now I'm lost. Can anyone explain to me how to do these type, step by step?

1. Express the complex number in polar form:

a) (e^i*pi/3)^2

b) (2+3i) + (-5-7i)

c) 4root(10e^i*pi/2)

d) (2i)^3 - (2i)^2 + 2i - 1

• Nov 23rd 2007, 04:14 AM
CaptainBlack
Quote:

Originally Posted by ohhhh
1. Express the complex number in polar form:

a) (e^i*pi/3)^2

$(e^{i \pi/3})^2=e^{i 2 \pi/3}$

which may be what you want of in $(r, \theta)$ form this is $(1, 2 \pi/3)$

RonL
• Nov 23rd 2007, 04:20 AM
CaptainBlack
Quote:

Originally Posted by ohhhh
1. Express the complex number in polar form:

b) (2+3i) + (-5-7i)

$
(2+3i) + (-5-7i) = -3 -4i = \sqrt{3^2+4^2}(\cos(\theta)+i\sin(\theta))=5e^{i\t heta}
$

where $\theta=-\arctan(4/3)$

RonL
• Nov 26th 2007, 09:58 AM
ohhhh
what about a problem like (-1-i)^4/3? how do i "find the quantity" and then change the form to cartesian?
• Nov 26th 2007, 08:30 PM
CaptainBlack
Quote:

Originally Posted by ohhhh
what about a problem like (-1-i)^4/3? how do i "find the quantity" and then change the form to cartesian?

$
-1-i =\sqrt{2} e^{\frac{3 \pi i}{2}+2 \pi i n}, \ n = 0, \pm1, ..
$

so:

$
(-1-i)^{4/3} =4^{1/3} e^{\frac{4 \pi i}{2}+8 \pi i n/3}=4^{1/3} e^{8 \pi i n/3}, \ n = 0, \pm1, ..
$

of which only three values are distinct:

$
(-1-i)^{4/3} =4^{1/3} e^{8 \pi i n/3}, \ n = 0, 1, 2
$

RonL