# Complex number question.

• May 24th 2010, 05:50 AM
scofield131
Complex number question.
Find the real and imaginary parts of (-2+2i)^20(1+isquareRt(3))

I don't really know how to solve this question, but I've tried and I haven't actually got a complex number. Could anybody tell me where I've gone wrong if it I've just done it the wrong way altogether. Sorry it looks quite messy.

(-2+2i)^20

sqroot(8) (cos(pi/4) + isin(pi/4)) -- found modulus and principle argument
sqroot(8)^20 (cos(20pi/4) + isin(20pi/4)) -- multiply modulus and principle argument to power of 20
sqroot(8)^20 (cos(pi) + isin(pi)) -- subtract principle argument by 360 until 0 < pi < 360

sqr(8)^20 (-1 + 0) = -(sqr(8)^20) -- find cos + sin values)

(1+iSqr(3))^12

2(cos(pi/3) + isin(pi/3))
4096(cos(12pi/3) + isin(12pi/3))
4096(cos(2pi) + isin(2pi))
4096(1 + 0)

4096 . -(sqr(8)^20) = -4.39 x 10^12?
• May 24th 2010, 06:09 AM
Prove It
Quote:

Originally Posted by scofield131
Find the real and imaginary parts of (-2+2i)^20(1+isquareRt(3))

I don't really know how to solve this question, but I've tried and I haven't actually got a complex number. Could anybody tell me where I've gone wrong if it I've just done it the wrong way altogether. Sorry it looks quite messy.

(-2+2i)^20

sqroot(8) (cos(pi/4) + isin(pi/4)) -- found modulus and principle argument
sqroot(8)^20 (cos(20pi/4) + isin(20pi/4)) -- multiply modulus and principle argument to power of 20
sqroot(8)^20 (cos(pi) + isin(pi)) -- subtract principle argument by 360 until 0 < pi < 360

sqr(8)^20 (-1 + 0) = -(sqr(8)^20) -- find cos + sin values)

(1+iSqr(3))^12

2(cos(pi/3) + isin(pi/3))
4096(cos(12pi/3) + isin(12pi/3))
4096(cos(2pi) + isin(2pi))
4096(1 + 0)

4096 . -(sqr(8)^20) = -4.39 x 10^12?

Convert them to polars:

$|-2 + 2i| = \sqrt{(-2)^2 + 2^2} = 2\sqrt{2}$

$\arg{(-2 + 2i)} = \pi - \arctan{\frac{2}{2}} = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

$|1 + \sqrt{3}i| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2$.

$\arg{(1 + \sqrt{3}i)} = \arctan{\frac{\sqrt{3}}{1}} = \frac{\pi}{3}$.

Therefore:

$(-2 + 2i)^{20}(1 + \sqrt{3}i) = \left(2\sqrt{2}e^{\frac{3\pi}{4}i}\right)^{20}\lef t(2e^{\frac{\pi}{3}i}\right)$

$= \left[\left(2\sqrt{2}\right)^{20}e^{15i}\right]\left(2e^{\frac{\pi}{3}i}\right)$

$= 2^{31}e^{15\pi i + \frac{\pi}{3}i}$

$= 2^{31}e^{\frac{46 \pi}{3}i}$

$= 2^{31}\left[\cos{\left(\frac{46\pi}{3}\right)} + i\sin{\left(\frac{46\pi}{3}\right)}\right]$

$= 2^{31}\left[\cos{\left(\frac{4\pi}{3}\right)} + i\sin{\left(\frac{4\pi}{3}\right)}\right]$

$= 2^{31}\left[-\cos{\left(\frac{\pi}{3}\right)} - i\sin{\left(\frac{\pi}{3}\right)}\right]$

$= 2^{31}\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)$

$= -2^{30} - 2^{30}\sqrt{3}i$.
• May 26th 2010, 10:35 AM
scofield131
Quote:

Originally Posted by Prove It
Convert them to polars:

$|-2 + 2i| = \sqrt{(-2)^2 + 2^2} = 2\sqrt{2}$

$\arg{(-2 + 2i)} = \pi - \arctan{\frac{2}{2}} = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

$|1 + \sqrt{3}i| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2$.

$\arg{(1 + \sqrt{3}i)} = \arctan{\frac{\sqrt{3}}{1}} = \frac{\pi}{3}$.

Therefore:

$(-2 + 2i)^{20}(1 + \sqrt{3}i) = \left(2\sqrt{2}e^{\frac{3\pi}{4}i}\right)^{20}\lef t(2e^{\frac{\pi}{3}i}\right)$

* $= \left[\left(2\sqrt{2}\right)^{20}e^{15i}\right]\left(2e^{\frac{\pi}{3}i}\right)$

$= 2^{31}e^{15\pi i + \frac{\pi}{3}i}$

$= 2^{31}e^{\frac{46 \pi}{3}i}$

$= 2^{31}\left[\cos{\left(\frac{46\pi}{3}\right)} + i\sin{\left(\frac{46\pi}{3}\right)}\right]$

$= 2^{31}\left[\cos{\left(\frac{4\pi}{3}\right)} + i\sin{\left(\frac{4\pi}{3}\right)}\right]$

$= 2^{31}\left[-\cos{\left(\frac{\pi}{3}\right)} - i\sin{\left(\frac{\pi}{3}\right)}\right]$

$= 2^{31}\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)$

$= -2^{30} - 2^{30}\sqrt{3}i$.

Hey, thanks for helping. However, could you explain it further from the star. I'm, in particular, unsure where the 15 came from?

Thanks.
• May 26th 2010, 11:02 AM
Plato
This is, I think, a simpler approach to this problem.
Because $\left( { - 2 + 2i} \right) = \sqrt 8 \exp \left( {\frac{{3\pi i}}{4}} \right)$ then $\left( { - 2 + 2i} \right)^{20} = \left( {\sqrt 8 } \right)^{20} \exp \left( {\frac{{3\pi i}}{4}} \right)^{20} = 8^{10} \exp \left( { 15\pi i} \right) = - 2^{30}$.
Now just multiply $-2^{30}$ by $(1+i\sqrt{3})$.
• May 26th 2010, 11:19 AM
scofield131
Quote:

Originally Posted by Plato
This is, I think, a simpler approach to this problem.
Because $\left( { - 2 + 2i} \right) = \sqrt 8 \exp \left( {\frac{{3\pi i}}{4}} \right)$ then $\left( { - 2 + 2i} \right)^{20} = \left( {\sqrt 8 } \right)^{20} \exp \left( {\frac{{3\pi i}}{4}} \right)^{20} = 8^{10} \exp \left( { 15\pi i} \right) = - 2^{30}$.
Now just multiply $-2^{30}$ by $(1+i\sqrt{3})$.

Thank you, that is simpler. However, call me stupid, I still don't understand why exp(3pi/4)^20 turned into 15pi? As when I do 135^20 I get 4.04 x 10^42, not 15pi?
• May 26th 2010, 11:32 AM
Plato
Quote:

Originally Posted by scofield131
Thank you, that is simpler. However, call me stupid, I still don't understand why exp(3pi/4)^20 turned into 15pi? As when I do 135^20 I get 4.04 x 10^42, not 15pi?

Recall that $\left[ {r\exp (\phi i)} \right]^n = r^n \exp (n\phi i)
$

$\left( {\frac{{3\pi i}}
{4}} \right)\left( {20} \right) = 15\pi i,\quad \left(\frac{3}
{4}\right)(20) = 15$
• May 26th 2010, 11:53 AM
scofield131
Quote:

Originally Posted by Plato
Recall that $\left[ {r\exp (\phi i)} \right]^n = r^n \exp (n\phi i)
$

$\left( {\frac{{3\pi i}}
{4}} \right)\left( {20} \right) = 15\pi i,\quad \left(\frac{3}
{4}\right)(20) = 15$

Thank you, I get it now. Sorry to keep bothering you, but could you perhaps explain the rest of the answer too? In particular, how you got to -2^30? And then if I then go back to the start and do the same for $\left( { 1 + i$ $sqrt{3}} \right)^{12} " alt="sqrt{3}} \right)^{12} " /> then multiply them together?
• May 26th 2010, 11:55 AM
scofield131
Quote:

Originally Posted by Plato
Recall that $\left[ {r\exp (\phi i)} \right]^n = r^n \exp (n\phi i)
$

$\left( {\frac{{3\pi i}}
{4}} \right)\left( {20} \right) = 15\pi i,\quad \left(\frac{3}
{4}\right)(20) = 15$

Thank you, I get it now. Sorry to keep bothering you, but could you perhaps explain the rest of the answer too? In particular, how you got to -2^30? And then if I then go back to the start and do the same for $\left( { 1 + i sqrt{3}} \right)^{12}$

Then multply them together?
• May 26th 2010, 11:56 AM
Plato
$\left( 8 \right)^{10}-\left( 2^3 \right)^{10}=2^{30}$