Mathematics: Post your doubts here!

[hellodjfos;s'ff]

Can you please explain this part again? I don't get the concept of 4^2 ?

Ok so we are trying to find out the radius of the (circle) complex number |z-4i| right? So how our earlier proof showed us that we need to square both the sides and gather the terms to show that one side of complex number is equal to the other. Basically, the proof zz*-2iz*+2iz-12=0 shows us that every complex number equation must boil down to this if both sides of equation are equal.
Let radius of the complex number=X so |z-4i|=X and now square both sides
zz*-2iz*+2iz+4-X^2=0
If you compare this to the equation earlier, both equations are almost identical with the difference being that the constant term is -12 in the other equation and in this equation it's 4-X^2.
We need to make the constant terms equal so that the equations are identical.
So -12=4-X^2... X will turn out to be 4 meaning that the radius is 4.

 

[hellodjfos;s'ff]

Can someone help me with Part (ii)
View attachment 64584
Oct/Nov/2016/61

For a blue pen to be in the left pocket once operation T takes place, a red pen must've been chosen from the left pocket followed by red pen from right pocket (1st) OR a blue pen chosen from left pocket and a blue pen chosen from right pocket(2nd).
For 1st scenario: (3/4)*(4/5)=0.6
2nd: (1/4)*(2/5)=0.1
0.6+0.1=0.7

 

[hellodjfos;s'ff]

Part b(ii)
View attachment 64585

You must find the combinations when only the batsman is being included in the squad(1st), only the bowler is being included in the squad(2nd) and when none of them are being included(3rd).
1st: 1C1(for the batsman having a confirmed place)*6C4*4C4(bowler has been excluded here)*2C1*2C1=60
2nd: 6C5*1C1(bowler having confirmed place)*4C3*2C1*2C1=96
3rd: 6C5(batsman excluded)*4C4(bowlers excluded)*2C1*2C1=24
24+96+60=180

 

[Hamnah Zahoor]

hellodjfos;s'ff Thank you for your help

 

[Hamnah Zahoor]

You must find the combinations when only the batsman is being included in the squad(1st), only the bowler is being included in the squad(2nd) and when none of them are being included(3rd).
1st: 1C1(for the batsman having a confirmed place)*6C4*4C4(bowler has been excluded here)*2C1*2C1=60
2nd: 6C5*1C1(bowler having confirmed place)*4C3*2C1*2C1=96
3rd: 6C5(batsman excluded)*4C4(bowlers excluded)*2C1*2C1=24
24+96+60=180

In the third why are we excluding both the batsman and the bowler.

 

[hellodjfos;s'ff]

In the third why are we excluding both the batsman and the bowler.

Because when we exclude both the batsmen and bowler, the batsman is still refusing to be in a team with a bowler but if they are not picked, they don't end up being in a team together anyways so the condition is met.

 

[Hamnah Zahoor]

Because when we exclude both the batsmen and bowler, the batsman is still refusing to be in a team with a bowler but if they are not picked, they don't end up being in a team together anyways so the condition is met.

Thank you

 

[A*****]

|z-10i|=2|z-4i|
(z-10i)(z+10i)=4(z-4i)(z+41) - using the concept from first equation in (i)
(z-10i)(z-10i)*=4(z-4i)(z-4i)* - now use concept from 2nd equation in (ii)
(z-10i)(z*+10i)=4(z-4i)(z*+4i)
zz*+10iz-10iz*+100=4(zz*+4iz-4iz*+16)
3zz*+6iz-6iz*-36=0
3(zz*-2iz*+2iz-12)=0 - first part proven
|z-2i|=(z-2i)(z*+2i)=zz*-2iz*+2iz+4
Now compare this with the earlier equation we have proven. If we subtract 16 from this new equation, both the equations would be equal right? So...
zz*-2iz*+2iz+4=4^2 - notice i out 4^2 not 16 because we were squaring both sides of the equation as the question had told us to do earlier.
Hence radius=4 making the equation |z-2i|=4

Can I work the 2nd part like this....
zz*-2iz*+2iz-12=0
zz*-2iz*+2iz+4-16=0
zz*-2iz*+2iz-4i²-16=0
z*(z-2i)+2i(z-2i)=4²
(z-2i)(z*+2i)=4²
(z-2i)(z-2i)*=4²
|z-2i|²=4²
|z-2i|=4
P.s can we take z-2i as whole in place of z because the statements in (i) are for z?

 

[hellodjfos;s'ff]

Can I work the 2nd part like this....
zz*-2iz*+2iz-12=0
zz*-2iz*+2iz+4-16=0
zz*-2iz*+2iz-4i²-16=0
z*(z-2i)+2i(z-2i)=4²
(z-2i)(z*+2i)=4²
(z-2i)(z-2i)*=4²
|z-2i|²=4²
|z-2i|=4
P.s can we take z-2i as whole in place of z because the statements in (i) are for z?

That works well too.

 

[A*****]

hellodjfos;s'ff in these questions how do we know that we have to take x in radians or degrees?

 

[A*****]

(iii) part plz...a bit confused about the minus sign in the argumemt of u

 

[hellodjfos;s'ff]

hellodjfos;s'ff in these questions how do we know that we have to take x in radians or degrees?

Usually they tell us whether to use radians or degrees. For complex number questions, try to use radians always (since that's the conventional rule) but they do accept answers in degrees. For a question like this where it's unclear whether to use radians or degrees, just look at how reasonable the trigonometric term is... so try out different values of x in sin(x/3) in degrees and radians mode and see which is more reasonable. Since most values of x produce a value too small in degree mode and most values of x produce a value large enough in radians, we do this question in radians.

 

[hellodjfos;s'ff]

(iii) part plz...a bit confused about the minus sign in the argumemt of u

arg(u*/u) means arg(u*)-arg(u)
so arg(u*/u)=arg(u*)-arg(u)
Find these arguments and you will be able to prove the equation.

 

[A*****]

arg(u*/u) means arg(u*)-arg(u)
so arg(u*/u)=arg(u*)-arg(u)
Find these arguments and you will be able to prove the equation.

u is 3-i and u* 3+i
The problem is when it would be written as arg(u*)-arg(u), which would be tan-¹(1/3)-tan-¹(-1/3), what would be done to the minus sign in arg(u)?

 

[hellodjfos;s'ff]

u is 3-i and u* 3+i
The problem is when it would be written as arg(u*)-arg(u), which would be tan-¹(1/3)-tan-¹(-1/3), what would be done to the minus sign in arg(u)?

-tan-¹(1/3) and tan-¹(-1/3) are the same so tan-¹(1/3)-tan-¹(-1/3) becomes tan-¹(1/3)+tan-¹(1/3) hence 2tan-¹(1/3)

 

[A*****]

-tan-¹(1/3) and tan-¹(-1/3) are the same so tan-¹(1/3)-tan-¹(-1/3) becomes tan-¹(1/3)+tan-¹(1/3) hence 2tan-¹(1/3)

Thank u sooo much!!!

 

[A*****]

-tan-¹(1/3) and tan-¹(-1/3) are the same so tan-¹(1/3)-tan-¹(-1/3) becomes tan-¹(1/3)+tan-¹(1/3) hence 2tan-¹(1/3)

Will tan-¹(1/-3) also equal -tan-¹(1/3)?

 

[hellodjfos;s'ff]

Will tan-¹(1/-3) also equal -tan-¹(1/3)?

Yes they're the same. You can even put the values in your calculator and clarify that.

 

[A*****]

Yes they're the same. You can even put the values in your calculator and clarify that.

So if a complex number is -5+2i so how will its argument be written in a presentable way? And it won't be equal to -tan-¹(2/5) right?

 

[hellodjfos;s'ff]

-5+2i would be in the 2nd quadrant so the argument would be pi/2+tan-¹(5/2)