# Use exponential notation to form a+ib

1. Mar 5, 2013

### thatguythere

1. The problem statement, all variables and given/known data
Use exponential notation to write (√3-i)(1+i√3) in the form a+ib.

2. Relevant equations

3. The attempt at a solution
Let (√3-i) = z1
r1=|z1|=√((√3)2-i2)
=√(3-1)
=√2

Therefore,
√3=√2cosθ and -1=√2sinθ
cosθ = √3/√2 and sinθ = -1/√2

θ= arcsin(-1/√2)
θ= -∏/4

z1 = √2(cos(-∏/4)+isin(-∏/4))

Let (1+i√3) = z2
r2 = |z2| = √(12+(i√3)2)
= √4
= 2

Therefore
1 = 2cosθ and 3 = 2sinθ
cosθ = 1/2 and sinθ = 3/2

However,
θ=arcsin(3/2) does not compute which leads me to believe that (i√3)2 = 3 is incorrect. Any help is appreciated. Thank you.

2. Mar 5, 2013

### SammyS

Staff Emeritus
What is i2 ?

3. Mar 5, 2013

### thatguythere

i2 = -1
However if I plug that in, I get this.
r2 = |z2| = √(12+(i√3)2}
= √(1+(-1(3))
= √-2
Which still does not compute.

4. Mar 5, 2013

### Dick

|a+bi|=sqrt(a^2+b^2).

5. Mar 5, 2013

### thatguythere

Ah, right. This leads me to another question. Do I use the "1" from i and eliminate the sqrt3, or treat it as (sqrt3)i and use the resulting 3? I would guess the latter.

6. Mar 5, 2013

### Dick

I'm not sure what you are asking. If z2=1+i*sqrt(3) then a=1 and b=sqrt(3). sqrt(a^2+b^2)=sqrt(1+3)=2.

7. Mar 5, 2013

### thatguythere

That is what I was asking and what I thought. Thank you. Now let's see if I can wrestle this into exponential form.

8. Mar 5, 2013

### Dick

Good. Your probably know using exponential form is a complete waste of time, right? You could just multiply (√3-i)(1+i√3) out and be done with it. But if that's what the ask you to do then that's what you should do. It'll be easy to check your answer.

9. Mar 5, 2013

### thatguythere

Well, the problem states "Use exponential notation to write (√3-i)(1+i√3) in the form a+ib." Have I been wasting my time?

10. Mar 5, 2013

### Dick

Not if the problem forces you to do it that way. I'm just saying you can easily check your answer by doing it the simple way.

11. Mar 5, 2013

### thatguythere

z1 = √2{cos(-∏/4)+isin(-∏/4)} = √2ei(-∏/4)

z2 = 2{cos∏/3+isin∏/3) = 2ei∏/3

z1z2 = 2√2ei(-∏/4)+i∏/3
=2√2ei∏/12

So far so good?

12. Mar 5, 2013

### Dick

z2=2*exp(i*pi/3), that's good. z1 isn't. You might be rehashing some of your old bad stuff into that one.

13. Mar 5, 2013

### thatguythere

z1 = 2{cos11∏/6+isin11∏/6}
= 2ei11∏/6

14. Mar 5, 2013

### Dick

I would have said 2exp(-i*pi/6), but sure, that's the same as 2exp(i*11pi/6). Think you are almost there.

15. Mar 5, 2013

### thatguythere

z1z2 = 2*2ei11∏/6+i∏/3
= 4ei13∏/6

4{cos13∏/6+isin13∏/6} Trig confuses me, so let's see.
4{-cos∏/6-isin∏/6}
4(-√3/2)-(i/2)
-2√3-2i

16. Mar 5, 2013

### Dick

4*exp(13*pi/6) is correct. I guess trig does confuse you. cos(13*pi/6) isn't equal to -cos(pi/6). Why would you think that?

17. Mar 5, 2013

### thatguythere

I don't know what I did there. Does it instead behave the same as ∏/6? Making my answer
2√3+2i?

18. Mar 5, 2013

### Dick

Sure, multiply out (√3-i)(1+i√3) to confirm that.

19. Mar 5, 2013

Thank you.