Question 1

How would you simplify (2i)(5+4i) ??

Accepted Answer

(2i)(5+4i)
i squared (i^2)= -1
So when you distribute you get 10i+ 8i^2
Since you know that i^=1, you can take 8(-1)= -8
So your final answer is: -8+10i

Question 2

I was wondering, Since addition and subtraction of complex numbers use traditional methods, why do complex number's require Binomial folding? not that I dislike folding it just seems a bit arbitrary at first.

Accepted Answer

Complex numbers use binomial methods of multiplication because unlike real numbers, imaginary numbers have two components.  Imaginary numbers are generally defined using the form a + bi where a and b are both real numbers.  Due to the fact that imaginary numbers have two parts (although a can be 0) we must multiply them using by either the distributive property or by FOILing.  Real numbers can also be defined by using the imaginary system of a + bi by merely having b equal to 0.

Question 3

What is Argand Plane in Complex Numbers? Is Khan covering it in any of the upcoming topics under complex numbers?

Accepted Answer

Thank you for your question.

I copied the following from the wikipedia article on complex numbers.  I had never heard the term Argand Plane prior to seeing your question.

"The complex plane is sometimes called the Argand plane because it is used in Argand diagrams. These are named after Jean-Robert Argand (1768–1822), although they were first described by Danish land surveyor and mathematician Caspar Wessel (1745–1818).[2] Argand diagrams are frequently used to plot the positions of the poles and zeroes of a function in the complex plane."

Yes you will see plots of complex numbers in the complex plane here in Khan academy, but I'm not sure that you will ever hear that term used.

Question 4

what is a complex number? describe in detail

Accepted Answer

A complex number is a number that can be written (that doesn't mean it is currently being expressed this way) as  a + bi, where a and b are real numbers.

At least in terms of numbers you will encounter at this level of study, this means that ALL numbers are complex numbers because all numbers can be expressed in the form of a + bi.

A real number is either a rational or irrational number. All real numbers can be expressed in terms of a + 0i. Thus, a real number is a special kind of complex number, specifically one in which the coefficient (b) of i is 0 when written in a+bi form.

A nonreal complex number (often just called a nonreal number) is a complex number which, when written in the form of a + bi, the b is NOT 0. As the name implies, all numbers that are not real numbers are nonreal complex numbers.

An imaginary number is a nonreal number which, when written in the a + bi form, the a IS 0. In other words, an imaginary number can be written as 0 + bi. This means that an imaginary number is a number that can be expressed is i times some real number other than 0.

NOTE: far too many teachers and textbooks get this wrong and call all nonreal numbers "imaginary numbers", so you need to check with your teacher to make sure you use the terms the way the teacher wants, even if it is wrong.
 
So, in summary, real numbers and imaginary numbers are both specical kinds of complex numbers:
`0 + bi` is the form of an imaginary number
`a + 0i` is the form of a real number.

Question 5

I don't get how we can say that i² = -1 ?

We don't know the value of i, if we did we would have solved the calculus much faster. Where does that come from ?

Accepted Answer

Hey! Well, actually in complex numbers i^2 is defined by rule as -1 and that is where we get the imaginary value of i. If you want to go the other way around, then you can simply square i although its value is imaginary: sqrt(-1)^2= -1.
In the first video on "The imaginary unit i" Sal talked about i and powers of i so you might want to check that out.

Hope this helped...

Question 6

what if you have something like (1-i)^3?

Accepted Answer

Expanding it we get
(1 - i) x (1 - i) x (1 - i)
The first two parentheses give us:
(1 -2i +i²) x (1 - i)
which simplifies . . .
 = (1 -2i -1) x (1 - i)
 =-2i x (1 - i)
Multiplying out gives us
-2i  -2i²
which simplifies. . . 
-2i + 2
= 2 - 2i

We can do a quick magnitude test to see if things are good:

To calculate the magnitude, take the root of the sum of the imaginary part squared and the real part squared. So for the original term (without the exponent):
(1 - i), the magnitude is square root( 1² + 1² ) = 2^(1/2)

The result was (2 - 2i), which has a magnitude of:
square root of (2² + 2²) = 8^(1/2) = 2^(3/2)

2^(3/2) is 2^(1/2)^3, so our answer (could) be correct.

Question 7

Can someone please explain the whole i^2 is equal to -1 thing? Also, why would -15^-1 be 15? Wouldn't be -1/15?

Accepted Answer

(-15)⁻¹ = 1/(-15) = -1/15
𝑖 is the imaginary unit. It opens up a whole new number system called the _imaginary numbers_. Try to think of a number that when multiplied by itself, gives a negative number. Well a positive number times itself is a positive number. So is a negative number times itself! So it may seem that no number can satisfy this property! In fact no _real_ number (a number in the real number system which is what you are used to working with), satisfies this property - only imaginary numbers do. This is because, we define:
𝑖 = √(-1)
Or equivalently:
𝑖² = -1
All imaginary numbers, when squared, give a negative real number. All imaginary numbers are multiples of 𝑖. The imaginary unit is analogous to 1 in the real number system - for all imaginary numbers are multiples of 𝑖. For example, 2𝑖, -44𝑖, π𝑖, or -𝑖√3. It may be hard to wrap your head around these numbers but they are just another way of thinking - just like when you were introduced to negative numbers (after all, what number can be less than nothing at all?). Comment if you have questions.

Question 8

Though my question sounds stupid, I want to know if we can use i as a variable.

Accepted Answer

i can be treated as a variable in the usual way, in that you can freely add, subtract, multiply, divide, take roots, and exponentiate with it. It's slightly nicer than a general variable, in fact, since we can freely divide by i (since we know i≠0) and we have the simplifying property that i²=-1.

If you're asking whether you can use the symbol i to stand in for an unknown, then you can do so. But to avoid confusion, I recommend against it if you're also going to be working with the imaginary unit i.

Question 9

how do you know that i squared is -1?

Accepted Answer

i = √(-1)
i^2 = √(-1) * √(-1) = √[(-1)(-1)]  = -1
because there are two of the same factors (-1) within the radical.

Question 10

How would I simplify (4+5i^3) + (-9 +11i)?

Accepted Answer

First, note that i³=-i. So your expression becomes
(4-5i)+(-9+11i)
We can also drop the parentheses, since addition is associative.
4-5i-9+11i
Now we just combine like terms, real and imaginary:
-5+6i

Integrated math 2

Course: Integrated math 2 > Unit 5

Multiplying complex numbers

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