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Mathematics: Post your doubts here!

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farhan141

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Fiery987 said:
Whoa, man! To find mod of a complex no., you use this formula:

If z = a+bi
then |z| = sqrt(a^2+b^2)

because if you plot z on an Argand diagram, you can find the length of the line (from origin to z) using Pythagoras theorem as it makes a triangle. And that length is mod of z. You getting it?

And what's iota? The 'i'? Well, i just differentiates between the real and unreal part of the complex no. so that is why it isn't written during calculations (e.g. when finding arg z, you don't write i in the formula)

Hope you understand it because this my first time explaining something on xtremepapers!
Click to expand...
thanks buddy. What abt part ii question i asked
 
Fiery987

Fiery987

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farhan141 said:
In second part i dont get it from 3rd step onwards. After squaring on both sides why is (z-2i)^2 equal to (z-2i)(z-2i)*. I dont get the conjuguate part (*)
Click to expand...

DON'T read the first part where I crossed it. The answer starts from |z-2i| = 4

Well, I used the relationship given in part (i). As they made us prove it, there must be some use of it, right? But I did it this way (as I proved in reverse) :

because |z|^2 = zz* {Part (i) question}
then |z-2i|^2 = (z-2i)(z-2i)* {z is a complex no. so z-2i must also be a complex no. because add/subt of a complx no. = another complx no.}
 
i_try9621

i_try9621

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oh okay.
So the question says "reflex" angle which means it's greater than 180 degree and less than 360 degree, which falls in 3rd and 4th quadrant where sin(theta) is negative. So you use a negative sign to indicate that.
 
qwertypoiu

qwertypoiu

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The complex one.
Note that for the diagram question, the origin of my circle should be at (0,2) but it looks like it's slightly higher. I only realised that later so yeah just imagine it like that :)
y61043tYKCSJ1WVaRffY305H.jpg
farhan141
 
qwertypoiu

qwertypoiu

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Fiery987 said:
Um, why did you eliminate x if x>0? x can't be 1 only.
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No I divided by x on both sides.
I justified this by reminding that x>0 so it cannot be equal to zero.
If you divide zero on both sides you can run into all sorts of trouble. But as long as x is not zero you can divide on both sides like that.
Or you can think of it as just factoring out x. Then the second bracket can be equated to zero as well.
 
Fiery987

Fiery987

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qwertypoiu said:
No I divided by x on both sides.
I justified this by reminding that x>0 so it cannot be equal to zero.
If you divide zero on both sides you can run into all sorts of trouble. But as long as x is not zero you can divide on both sides like that.
Or you can think of it as just factoring out x. Then the second bracket can be equated to zero as well.
Click to expand...

Ohhh, right. Thanks a lot!
 
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