Use exponential notation to form a+ib

[thatguythere]

Homework Statement

Use exponential notation to write (√3-i)(1+i√3) in the form a+ib.

Homework Equations

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)
= √(1+(1(3)) <-- I am very uncertain about this step
= √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.

 

[SammyS]

Homework Statement

Use exponential notation to write (√3-i)(1+i√3) in the form a+ib.

Homework Equations

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)
= √(1+(1(3)) <-- I am very uncertain about this step
= √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.

What is i² ?

 

[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.

 

[Dick]

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.

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

 

[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.

 

[Dick]

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.

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.

 

[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.

 

[Dick]

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

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.

 

[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?

 

[Dick]

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?

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.

 

[thatguythere]

z₁ = √2{cos(-∏/4)+isin(-∏/4)} = √2e^i(-∏/4)

z₂ = 2{cos∏/3+isin∏/3) = 2e^i∏/3

z₁z₂ = 2√2e^{i(-∏/4)+i∏/3}
=2√2e^i∏/12

So far so good?

 

[Dick]

z₁ = √2{cos(-∏/4)+isin(-∏/4)} = √2e^i(-∏/4)

z₂ = 2{cos∏/3+isin∏/3) = 2e^i∏/3

z₁z₂ = 2√2e^{i(-∏/4)+i∏/3}
=2√2e^i∏/12

So far so good?

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.

 

[thatguythere]

z1 = 2{cos11∏/6+isin11∏/6}
= 2e^i11∏/6

 

[Dick]

z1 = 2{cos11∏/6+isin11∏/6}
= 2e^i11∏/6

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

 

[thatguythere]

z1z2 = 2*2e^{i11∏/6+i∏/3}
= 4e^i13∏/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

 

[Dick]

z1z2 = 2*2e^{i11∏/6+i∏/3}
= 4e^i13∏/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

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?

 

[thatguythere]

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

 

[Dick]

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

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

 

[thatguythere]

Thank you.