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## Algebra (all content)

### Course: Algebra (all content) > Unit 16

Lesson 11: Challenging complex number problems# Challenging complex numbers problem (3 of 3)

Paper 1 Problem 39 from the challenging 2010 IIT JEE exam, about complex numbers. Created by Sal Khan.

## Want to join the conversation?

- I don't understand how it made the demand C, why are the numbers one above the other?

-Excuse my English and thanks(13 votes)- Demand C is a Matrix. That is why the numbers are one above the other. You should look for the Matrices videos and the Determinant video so as to understand the option C and the way to solve it.(4 votes)

- Could someone explain why he cross multiplied and subtracted equations?(5 votes)
- Matrices are hard to express in this note editor, which is why people put the links in their answers. Anyway,

Let the first row be

|a b|

Let the second row be

|c d|

So your matrix**M**is

|a b|

|c d|

The determinant of**M**is found by multiplying a times d and subtracting the product of c and b

In other words, the determinant = (ad) - (cb)

If your matrix is

|2 1|

|2 -3|

The determinant is (2∙ -3) - (2∙1) = -6 -2 = -8(11 votes)

- Isn't there a shortcut to get the same solutions?(4 votes)
- just substitute some values, say, t=1/2, z1=2, z2=4, so we get z3= 3, and we get the answers in not more than 1 min..... :- )(9 votes)

- Is there a faster method to do the problem as in the exam we dont much time to go over each and every option?(4 votes)
- What I did was solve for Z, Z1 and Z2 and I got all of the answers correct. It was a lot quicker.(3 votes)

- At1:10, Sal explains that he knows that the conjugate of z minus the conjugate of zsub1 equals the difference of zsub and zsub1 times t. He says that by looking at the negative imaginary parts you can trace that this is true, but when he did it at first to solve for z - zsub1, he used the provided definition for z. At this point, how can you just switch signs on the imaginary part and know what to do in terms of the given definition, which has zsub2 and zsub1 - I can't see how switching the imaginary part tells me how to change the signs on that other side, much less the whole other side of z - zsub1...(2 votes)
- Oh, well, that is because
`the conjugate`

of a complex number is merely the same number with a different sign on the**imaginary part**of the complex number. Sorry if that goes in a circle--it is the actual**definition**of conjugate. He repeated the definition about 5 times there as he talked about how you could go ahead and confirm what he is saying.

Example

A complex number of the general form`a + bi`

such as

3 + 4i

This COULD be used for z2, for example, at least for guess and check purposes.

a = 3

b = 4

The conjugate of`3 + 4i`

is`3 - 4i`

Another complex number

-2 + 8i

You could use that as z1, I suppose

a = -2

b = 8

The conjugate of`-2 + 8i`

is`-2 - 8i`

Notice that**only**the sign of the imaginary part of the complex number is changed

Another complex number

6 - 2i

a = 6

b = -2

The conjugate of`6 - 2i`

is`6 + 2i`

All he is saying is that, if you go back and substitute`a + bi`

for Z and`a1 + b1i`

for z1, and so on, AND for all the conjugates like zbar (you would have`a - bi`

for zbar--notice that "the sign of the imaginary part of the complex number has been swapped")

then you could go back to the first video in this series and re-prove (with about a half day of substituting) that

z- z1 = t(z2 - z1) using the general expansion I just gave and eventually that this determinant is also truly equal to zero.(6 votes)

- C) is a matrix ,but why is he assuming determinant?

Option C doesnt bring determinant at all it just shows matrix? why?(3 votes) - Sir,if (sinx/2+cosx/2+itanx)/(1+2isinx/2) is purely real, then all possible values of x are?

I'm finding difficulty in solving this plz help!

Thanks(1 vote)- Tarandeep just multiply the numerator and denominator by 1-2isinx/2

After that take the part with i and equate it to 0(3 votes)

- I don't understand at @1:20that how does conjugate of

conjugate Z - conjugate Z(sub1) = t ( conjugate(Z(sub2)) - conjugate(Z(sub1)) ).(2 votes)- p - u = tw

p, u, and w are complex numbers, t is a scalar

Let's define a function cge(z) = conjugate(z), where z is a complex number.

There's a pattern or rule you can follow which can be derived from observations or testing out different values to see this pattern. The pattern can be stated as:

The conjugate of a difference of complex numbers is equal to the difference of the conjugate of each of those complex numbers.

In other words, cge(p - u) = cge(p) - cge(u)

With this we can then show that cge(z) - cge(z1) = t(cge(z2) - cgez1))

Now with the above rule we can definitely see the relationship.

p - u = tw

Take the conjugate of both sides

cge(p - u) = cge(tw)

Split up the conjugate of the difference into the difference of two conjugates (our rule).

cge(p) - cge(u) = cge(tw)

The conjugate of a scalar is still a scalar

cge(p) - cge(u) = t*cge(w)

We can apply the above steps to our original expression

z - z1 = t(z2 - z1)

cge(z - z1) = cge(t(z2 - z1))

cge(z) - cge(z1) = t*cge(z2 - z1)

cge(z) - cge(z1) = t*(cge(z2) - cge(z1))

And there you have it.

A little caveat, I don't even know if the rule is actually a rule. It was just a observation that I made after testing out different complex numbers.(2 votes)

- Answer D is just a matrix. It never said anything about its determinant, so why is that correct?(1 vote)
- Actually, the notation is that a matrix is enclosed in brackets ([ ]). But option C is enclosed in bars (| |). When you have an array of numbers enclosed in bars, it's referring to the determinant of the matrix.(1 vote)

- i didn't understand why he found the determinent.(1 vote)

## Video transcript

So we're in the home stretch. Let's see if part C is true, and
I'm running out of real estate. So let me copy and
paste it actually. Actually, maybe I could do
it over here to the left. I had some doodles here. Let me clear those out. So let's see if we can do
it over here to the left. Part C. So we have
this determinant, and it's claiming
that it equals 0, so we have to see
if that's true. And so we can actually try
to multiply these out and see if anything interesting happens,
but we could even better leverage some of the
algebra that we've already done for part A. In part A, we
figured out that z minus z1 is equal to t times z2 minus z1. So this thing right
over here, this is the same thing as
t times z2 minus z1, and the reason why
that's interesting is I have a z1 minus
z1 here, so I'm starting to have similar things. And then this thing
over here-- so if I take the conjugate of each
of these, the conjugate of z minus the conjugate
of z1, this is going to be equal to t
times the conjugate of z2 minus the conjugate of z1. And I know that
because when I do this, I'm swapping all of the
signs on the imaginary parts of the complex numbers. So if you were to go through
this entire process just swapping the imaginary
parts of the complex number, the end result would have
it's imaginary parts swapped, and that's exactly
what we have over here. We have the imaginary
parts swapped over here. So if you want to work it out,
set this equal to A plus BI or set this equal to A1 plus
B1, you can work it out, but I think it's a pretty
intuitive idea that we're just making all of the-- we're
just swapping the signs on all of the imaginary parts of
each of these complex numbers. Now, with this said,
this determinant becomes pretty simple. This becomes t
times z2 minus z1. This becomes t times the
conjugate of z2 minus z1. This down here is z2 minus z1. And then this over here is
the conjugate of z2 minus z1. And so what is this determinant? It's going to be
this times this. So it's t times z2
minus z1 times z2 minus z1 times the
conjugate of those. So it's that times that
minus this times this, so minus t z2 z1, z2 minus
z1 times the conjugate of z2 minus the conjugate of z1. Now, this is exactly
equal to this. These things are obviously
going to cancel out, and we're clearly
going to get 0. So C is also true. So the correct answers
to this original problem were A, C, and D.