Complex number question.

Nov 2009
25
0
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?
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
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:

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

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


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

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


Therefore:

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

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

\(\displaystyle = 2^{31}e^{15\pi i + \frac{\pi}{3}i}\)

\(\displaystyle = 2^{31}e^{\frac{46 \pi}{3}i}\)

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

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

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

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

\(\displaystyle = -2^{30} - 2^{30}\sqrt{3}i\).
 
  • Like
Reactions: scofield131
Nov 2009
25
0
Convert them to polars:

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

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


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

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


Therefore:

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

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

\(\displaystyle = 2^{31}e^{15\pi i + \frac{\pi}{3}i}\)

\(\displaystyle = 2^{31}e^{\frac{46 \pi}{3}i}\)

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

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

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

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

\(\displaystyle = -2^{30} - 2^{30}\sqrt{3}i\).
Hey, thanks for helping. However, could you explain it further from the star. I'm, in particular, unsure where the 15 came from?

Thanks.
 

Plato

MHF Helper
Aug 2006
22,458
8,632
This is, I think, a simpler approach to this problem.
Because \(\displaystyle \left( { - 2 + 2i} \right) = \sqrt 8 \exp \left( {\frac{{3\pi i}}{4}} \right)\) then \(\displaystyle \left( { - 2 + 2i} \right)^{20} = \left( {\sqrt 8 } \right)^{20} \exp \left( {\frac{{3\pi i}}{4}} \right)^{20} = 8^{10} \exp \left( { 15\pi i} \right) = - 2^{30} \).
Now just multiply \(\displaystyle -2^{30}\) by \(\displaystyle (1+i\sqrt{3})\).
 
  • Like
Reactions: scofield131
Nov 2009
25
0
This is, I think, a simpler approach to this problem.
Because \(\displaystyle \left( { - 2 + 2i} \right) = \sqrt 8 \exp \left( {\frac{{3\pi i}}{4}} \right)\) then \(\displaystyle \left( { - 2 + 2i} \right)^{20} = \left( {\sqrt 8 } \right)^{20} \exp \left( {\frac{{3\pi i}}{4}} \right)^{20} = 8^{10} \exp \left( { 15\pi i} \right) = - 2^{30} \).
Now just multiply \(\displaystyle -2^{30}\) by \(\displaystyle (1+i\sqrt{3})\).
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?
 

Plato

MHF Helper
Aug 2006
22,458
8,632
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 \(\displaystyle \left[ {r\exp (\phi i)} \right]^n = r^n \exp (n\phi i)
\)

\(\displaystyle \left( {\frac{{3\pi i}}
{4}} \right)\left( {20} \right) = 15\pi i,\quad \left(\frac{3}
{4}\right)(20) = 15\)
 
  • Like
Reactions: scofield131
Nov 2009
25
0
Recall that \(\displaystyle \left[ {r\exp (\phi i)} \right]^n = r^n \exp (n\phi i)
\)

\(\displaystyle \left( {\frac{{3\pi i}}
{4}} \right)\left( {20} \right) = 15\pi i,\quad \left(\frac{3}
{4}\right)(20) = 15\)
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 \(\displaystyle \left( { 1 + i\)\(\displaystyle sqrt{3}} \right)^{12} \) then multiply them together?
 
Nov 2009
25
0
Recall that \(\displaystyle \left[ {r\exp (\phi i)} \right]^n = r^n \exp (n\phi i)
\)

\(\displaystyle \left( {\frac{{3\pi i}}
{4}} \right)\left( {20} \right) = 15\pi i,\quad \left(\frac{3}
{4}\right)(20) = 15\)
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 \(\displaystyle \left( { 1 + i sqrt{3}} \right)^{12} \)

Then multply them together?
 

Plato

MHF Helper
Aug 2006
22,458
8,632
\(\displaystyle \left( 8 \right)^{10}-\left( 2^3 \right)^{10}=2^{30}\)