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Mathematics: Post your doubts here!

asd

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can someone please help me with the hence show part in part (ii)??View attachment 39317
|z| = √[(a)^2+(b)^2], and |z|^2 = a^2+b^2
z.z*= (a+ib)(a-ib)= a^2 +aib -aib -(bi)^2 = a^2 + b^2
(N.B i^2 = -1)

(z-ki) = a+ib-ki = a +(b-k)i
(z-ki)*= a -(b-k)i = a-bi +ki
Ooops, HAHAHA never saw you wrote part (ii) :p
can't do that one either, I guess Namehere can help you with this :p
 
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in question it is 3+x^2 but u have multiplied with x^2...?? shouldn't it be...

(3+[sqrt3 tan(thita)]^2)^-1/2

this was the question.. no?
 

asd

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I've solved this but the working is pretty lengthy.
Ill just tell you in short, you gotta break the 1/x^2(2x+1) into partial fractions, and then integrate. youre gonna get -2ln(x) -1/x + 2ln(2x+1)= lny +c
Then use (1,1) to find c. Then youre gonna get too many ln's. just put them together in a single ln function (divide the ones being subtracted, multiply the ones being added) and then cancel the ln from both sides. THEN, substitute 2 in x and youre gonna get 25e/36e^(1/2) = 25e^(1/2) / 36
 
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|z| = √[(a)^2+(b)^2], and |z|^2 = a^2+b^2
z.z*= (a+ib)(a-ib)= a^2 +aib -aib -(bi)^2 = a^2 + b^2
(N.B i^2 = -1)

(z-ki) = a+ib-ki = a +(b-k)i
(z-ki)*= a -(b-k)i = a-bi +ki
Ooops, HAHAHA never saw you wrote part (ii) :p
can't do that one either, I guess Namehere can help you with this :p

I´m actually looking for someone who can solve Q7(ii), the part of hence show that the modulus of z-2i = 4.
Here is the question once again:
 

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