Question

How to expressing something in a+bi form?

http://prntscr.com/b2m7f1

Answer 1

First figure out what's (1 + i)^2

Answer 2

It is... a complex number?

Answer 3

I would first convert to trig form

z = r*(cos(theta) + i*sin(theta))

then use De Moivre's Theorem

Answer 4

But don't I have to square i^20?

Answer 5

for the number z = a+bi, we can find r and theta by using these formulas

\[\Large r = \sqrt{a^2+b^2}\]
\[\Large \theta = \arctan\left(\frac{b}{a}\right)\]

Note about theta: if a+bi is in Q2 or Q3, then you'll have to add on pi radians to theta

Answer 6

@sloppycanada you're given (1+i)^20 not just i^20

Answer 7

z = 1+i
z = 1+1*i
the number is in the form z = a+b*i where

a = 1
b = 1

Answer 8

So i converted it. I got 2(cos pi/2)) + isin(pi/2)

Answer 9

you should get

r = sqrt(2)
theta = pi/4

try again

Answer 10

Here is another way:

\(\Large (1 + i)^{20} =\)

\(\Large = \left( (1 + i)^2 \right)^{10}\)

\(\Large =(1 + 2i - 1)^{10}\)

\(\Large =(2i)^{10} \)

\(\Large = 2^{10} \times i^{10} \)

\(\Large =1024 \times (-1) \)

\(\Large = -1024\)

Answer 11

\(\Large -1024 = -1024 + 0i\)

Answer 12

I don't understand how I get b/a with i^20?

Answer 13

But assuming I just i^20/1^20, I do get pi/4

Answer 14

@mathstudent55 has a great way to do it

here is how to do it using De Moivre's Theorem

\[\Large z = 1+1i\]

\[\Large z = \sqrt{2}*\left(\cos\left(\frac{\pi}{4}\right)+i*\sin\left(\frac{\pi}{4}\right)\right)\]

\[\Large z^{20} = \left[\sqrt{2}*\left(\cos\left(\frac{\pi}{4}\right)+i*\sin\left(\frac{\pi}{4}\right)\right)\right]^{20}\]

\[\Large z^{20} = \left(\sqrt{2}\right)^{20}*\left(\cos\left(20*\frac{\pi}{4}\right)+i*\sin\left(20*\frac{\pi}{4}\right)\right)\]

\[\Large z^{20} = \left(\sqrt{2}\right)^{20}*\left(\cos\left(5\pi\right)+i*\sin\left(5\pi\right)\right)\]

\[\Large z^{20} = 1024*\left(\cos\left(5\pi\right)+i*\sin\left(5\pi\right)\right)\]

\[\Large z^{20} = 1024*\left(-1+i*0\right)\]

\[\Large z^{20} = -1024\]

Answer 15

a = 1
b = 1

-------------------------------

\[\Large r = \sqrt{a^2+b^2}\]
\[\Large r = \sqrt{1^2+1^2}\]
\[\Large r = \sqrt{1+1}\]
\[\Large r = \sqrt{2}\]

-------------------------------

\[\Large \theta = \arctan\left(\frac{b}{a}\right)\]
\[\Large \theta = \arctan\left(\frac{1}{1}\right)\]
\[\Large \theta = \arctan\left(1\right)\]
\[\Large \theta = \frac{\pi}{4}\]

Answer 16

@jim_thompson5910 But what mathstudent did makes no sense to me. I get that he broke down the ^20, but I don't get where that -1 came from.

Answer 17

you mean how did @mathstudent55 turn `i^10` into `-1` ?

Answer 18

or some other step?

Answer 19

@sloppycanada

Answer 20

Oh, right, i^2 turns into -1, doesn't it?

Answer 21

It's the 1024 x -1 that screws me up. I don't understand where that -1 suddenly came from.

Answer 22

i^(10) = i^(2*5)
i^(10) = (i^2)^5
i^(10) = (-1)^5
i^(10) = -1

does that help clear things up?

Answer 23

`Oh, right, i^2 turns into -1, doesn't it?`

yes by definition
\[\Large i = \sqrt{-1}\]
so squaring both sides leads to
\[\Large i^2 = -1\]

Answer 24

It comes down to when I try to replicate it with a problem like this http://prntscr.com/b2mfwn

my answer falls apart.

Answer 25

What is (1-i)^2 equal to?

Answer 26

1 and -1

Answer 27

(1-i)^2 = (1-i)*(1-i)

now FOIL it out or use the box method. I prefer the box method