Complex numbers (1 Viewer)

[Antwan23q]

|z-8|=2Re(z-2)

sketch the locus and write equation.
its probably something really simple but it ive probably forgotten, i thought it was a circle lol

 

[shafqat]

|z-8|=2Re(z-2)

sketch the locus and write equation.
its probably something really simple but it ive probably forgotten, i thought it was a circle lol

Let z = x + iy
|x -8 + iy| = 2(x - 2)
(x-8)^2 - y^2 = 4(x-2)^2

Sorry, my mistake. Fixed now. Thanks for completing solution.

 

[011]

Let z = x + iy
|x -8 + iy| = 2(x - 2)
(x-8)^2 + y^2 = 4(x-2)^2

((x-8) + iy)^2 = (x-8)^2 - y^2? (equating re)

x²-16x+64 - y² = 4(x²-4x+4)
x²-16x+64 - y² = 4x²-16x+16
3x²+ y² = 48

 

[Tuna]

also can anyone help me with this one:

simplify : (1+3^1/2) + i(1-3^1/2)/(1+i.3^1.2) how do you simplify that

 

[yrtherenonames]

"also can anyone help me with this one:

simplify : (1+3^1/2) + i(1-3^1/2)/(1+i.3^1.2) how do you simplify that"

Assuming you meant, in the last part, i.3^1.5; you just realise the denominator.
If it really is i.3^1.2,,,,,,I don't think they would ask that.

 

[Tuna]

in polar form

 

[Tuna]

correction [(1+3^1/2) + i(1-3^1/2)]/(1+i.3^1.2)

 

[011]

correction [(1+3^1/2) + i(1-3^1/2)]/(1+i.3^1.2)

Do you mean 1.2 as in 6/5?
Like for the denominator 1+ i(3^(6/5))?

 

[Trev]

correction [(1+3^1/2) + i(1-3^1/2)]/(1+i.3^1.2)

=[(1+√3) + i(1-√3)](1-i√3)]/[(1-i√3)(1+i√3)]
=[1-i√3+√3-3i+i+√3-i√3-3]/[1+3]
= 1/2[-1-i-i√3+√3)
=1/2[(-1+√3)+i(-1-√3)]
Maybe a mistake there in the process, but it's just rationalising the denominator then an algebra bash.
EDIT: Had it as ^1/2 not ^1.2.... my bad

 

[FD3S-R]

can someone help:

cos2Θ - isin2Θ expressed in the form (cisΘ)ⁿ

 

[haboozin]

Sorry, my mistake. Fixed now. Thanks for completing solution.

woah u made a mistake.

 

[011]

can someone help:

cos2Θ - isin2Θ expressed in the form (cisΘ)ⁿ

cos2Θ - isin2Θ = cos2Θ + isin(-2Θ), since sinΘ is odd
= cos(-2Θ) + isin(-2Θ), since cosΘ is even
= cis(-2Θ)
= (cisΘ)^(-2), n = -2

 

[FD3S-R]

i understand u moving the minus infront of theta on the sin side, but how come there u put a minus also infront of the cos??

 

[KFunk]

i understand u moving the minus infront of theta on the sin side, but how come there u put a minus also infront of the cos??

Cos is an even function so:

F(x) = cosx ---> F(x) = F(-x)

so cos(2Θ ) = cos(-2Θ )

 

[FD3S-R]

damn thanks alot

 

[011]

i understand u moving the minus infront of theta on the sin side, but how come there u put a minus also infront of the cos??

cos is an even function, ie its graph is symmetrical in the y-axis. This of course means that you will get the same value of y for any theta/minus-theta combination. For example, 60 degrees will give you 1/2, and since the graph is symmetrical, -60 degrees will also give you 1/2 (same output of y).

So in more broad theoretical terms, cosx = cos(-x).

 

[FD3S-R]

sorry another question:

1-2i is one root of the equation x^2 +(1+i)x+k=0. Find the other root and the value of k.

i can find the value of k which makes the equation x^2+(1+i)x+5i=0

but i cant find the other root, ive attempted long division but cant seem to get it, can anyone solve without inspection or guessing thanks

 

[shafqat]

sorry another question:

1-2i is one root of the equation x^2 +(1+i)x+k=0. Find the other root and the value of k.

i can find the value of k which makes the equation x^2+(1+i)x+5i=0

but i cant find the other root, ive attempted long division but cant seem to get it, can anyone solve without inspection or guessing thanks

Try sum of roots.

 

[011]

sorry another question:

1-2i is one root of the equation x^2 +(1+i)x+k=0. Find the other root and the value of k.

i can find the value of k which makes the equation x^2+(1+i)x+5i=0

but i cant find the other root, ive attempted long division but cant seem to get it, can anyone solve without inspection or guessing thanks

shafqat is correct.
Let the other root be a+bi.

1 - 2i + a + bi = -1-i
(1 +a) + (b- 2)i = -1-i
Equate reals: a=-2
Equate imaginary: b=1
So the second root is -2+i
For k, do product of roots: k = (1-2i)(-2+i)

 

[velox]

haha i could actually remember how to do that question.