Complex Number Question (Easy)

[David Donald]

Homework Statement

Verify that i²=-1
using
(a+bi)(c+di) = (ac-bd)(ad+bc)i

Homework Equations

(a+bi)(c+di) = (ac-bd)(ad+bc)i

The Attempt at a Solution

I tried choosing coefficients so that it would be (i)(i) = (0 - 1)+(0+0)i = -1
so then I get i^2 = -1

But I was told that this was wrong and to try again...

Can anyone explain what I did was wrong or if
theirs a smarter way to verify?

 

[member 587159]

Homework Statement

Verify that i²=-1
using
(a+bi)(c+di) = (ac-bd)(ad+bc)i

Homework Equations

(a+bi)(c+di) = (ac-bd)(ad+bc)i

The Attempt at a Solution

I tried choosing coefficients so that it would be (i)(i) = (0 - 1)+(0+0)i = -1
so then I get i^2 = -1

But I was told that this was wrong and to try again...

Can anyone explain what I did was wrong or if
theirs a smarter way to verify?

We define i as the root of -1. So $$i^2 = -1$$ is true, no need to verify this.

The following statement is false: $$(a+bi)(c+di) = (ac-bd)(ad+bc)i$$
This would mean that the product of 2 complex numbers is an imaginary number. This is false. For example, $$i * i = i^2 = -1$$ is a real number.

What is $$(a+bi)(c+di)$$ equal to, using distributivity and $$i^2 = -1$$

 

[Mark44]

Homework Statement

Verify that i²=-1
using
(a+bi)(c+di) = (ac-bd)(ad+bc)i

Are you sure that you have written the equation above correctly? As already noted by Math_QED, this is false.

 

[PeroK]

It's supposed to be:

$(a+bi)(c+di) = (ac-bd) + (ad+bc)i$

And, assuming this holds for all $a, b, c, d$ show that $i^2 = -1$.

That's what I assume the exercise to be.

 

[Mark44]

We define i as the root of -1. So $i^2 = -1$ is true, no need to verify this.

But the whole point of this exercise is to prove this. In other words, the OP can't use this definition.

All,
Let's sit back and see what the OP says...