Imaginary number problem

[Acklebee]

I am stumped about how to proceed with this.
(1+i) to the fifth power.

 

[srmichael]

I am stumped about how to proceed with this.
(1+i) to the fifth power.

Do you know de Moivre's formula?

 

[pka]

I am stumped about how to proceed with this.
(1+i) to the fifth power.

\(\displaystyle 1 + i = \sqrt 2 \left[ {\cos \left( {\frac{\pi }{4}} \right) + i\sin \left( {\frac{\pi }{4}} \right)} \right]\)

Use de Moivre's theorem

 

[Acklebee]

No!

Do you know de Moivre's formula?

Did I write the question properly?
Here it is again -
(1 + i )^5

Does this make a difference?

 

[Bob Brown MSEE]

= (1 + i )(1 + i )(1 + i )(1 + i )(1 + i )
= (2i)(2i)(1 + i )
= -4(1 + i )

 

[pka]

Did I write the question properly?
Here it is again -
(1 + i )^5
Does this make a difference?

Do you know de Moivre's theorem?

If not expand it by hand. Look at this.

\(\displaystyle a=1~\&~b=i\)

 

[HallsofIvy]

I am stumped about how to proceed with this.
(1+i) to the fifth power.

Are you saying you do not know how to multiply complex numbers?
\(\displaystyle (1+ i)^2= 1+ 2i+ i^2= 1+ 2i- 1= 2i\)
\(\displaystyle (1+ i)^4= [(1+ i)^2]^2= [2i]^2= 4i^2= -4\)
To find \(\displaystyle (1+ i)^5\) multiply one more time by (1+ i): -4(1+ i)= -4- 4i.

 

[Acklebee]

= (1 + i )(1 + i )(1 + i )(1 + i )(1 + i )
= (2i)(2i)(1 + i )
= -4(1 + i )

Perfect - I can follow this! Thanks a lot!!!!

 

[Acklebee]

Are you saying you do not know how to multiply complex numbers?
\(\displaystyle (1+ i)^2= 1+ 2i+ i^2= 1+ 2i- 1= 2i\)
\(\displaystyle (1+ i)^4= [(1+ i)^2]^2= [2i]^2= 4i^2= -4\)
To find \(\displaystyle (1+ i)^5\) multiply one more time by (1+ i): -4(1+ i)= -4- 4i.

This helps a lot - thanks for going to the trouble of working this out!!! Spectacular!