# Another Complex Number Question... (1 Viewer)

thanks all!!
xx

#### ohexploitable

##### hey
$\bg_white \\\text{34) i. }x+iy+\frac{1}{x+iy}\\=x+iy+\frac{x-iy}{x^2+y^2}\\=\frac{x(x^2+y^2)+x}{x^2+y^2}+\frac{(y(x^2+y^2)-y)i}{x^2+y^2}\\=\frac{x(x^2+y^2+1)}{x^2+y^2}+\frac{y(x^2+y^2-1)i}{x^2+y^2}$

#### Bored Of Fail 2

##### Banned
finding th modulus for 35 is fairly standard, just mod = sqrt ( (real part)^2 + (imag part)^2 )

= sqrt ( ( 1+cosx)^2 + (sin(x))^2 ) = sqrt ( 2 +2cosx )

then using double angle results cosx = 2 (cos(x/2) )^2 -1 , then sub in and mod is done

argument is a fair bit harder

#### ohexploitable

##### hey
$\bg_white \\\text{35) i. }r\text{cis}\theta=1+\cos\theta+i\sin\theta\\r=\sqrt{1+2\cos\theta+\cos^2\theta+\sin^2\theta}\\r=\sqrt{2+2\cos\theta}\\r=\sqrt{2+4\cos^2\frac{\theta}{2}-2}\\r=2\cos\frac{\theta}{2}$

#### ohexploitable

##### hey
$\bg_white \\\tan\frac{\theta}{2}\\=\frac{\sin\theta/2}{\cos\theta/2}\\=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}\\=\sqrt{\frac{1-\cos^2\theta}{(1+\cos\theta)^2}}\\=\frac{\sin\theta}{1+\cos\theta}$

$\bg_white \\\text{arg}(z)=\tan^{-1}\frac{\sin\theta}{1+\cos\theta}\\=\tan^{-1}\left(\tan\frac{\theta}{2}\right)\\=\frac{\theta}{2}$

#### ohexploitable

##### hey
$\bg_white \\\text{For }(1+\cos\theta+i\sin\theta)^n,\\r=2^n\cos^n\frac{\theta}{2}\\\text{arg}(z)=\frac{n\theta}{2}$

#### Bored Of Fail 2

##### Banned
when you go from sin (x/2) / cos ( x/2) = sqrt ( (1-cosx)/ ( 1+cosx) ) I think I kinda know what did , but im thinking there is 90% chance that the person that asked the question has no idea what you did lol

Also, I must say I am very impressed someone doing gen maths could do these questions

#### ohexploitable

##### hey
$\bg_white \\\cos\theta=2\cos^2\frac{\theta}{2}-1\\\cos\frac{\theta}{2}=\sqrt{\frac{1+\cos\theta}{2}}\\\\\\\\\cos^2\frac{\theta}{2}=\frac{1+\cos\theta}{2}\\1-sin^2\frac{\theta}{2}=\frac{1+\cos\theta}{2}\\\sin\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{2}}$

#### Bored Of Fail 2

##### Banned
ahh I gotcha, and the third part to that questions is relatvely easy,

just a fairly standard binomial question where you take n=4

ie consider ( 1 + ( cosx+ i sinx ) ) ^4 and then expand it using binomial theorm and use the theorm ( cos(x) + isin(x) ) ^n = cos(nx) + i sin(nx)

then equate real and imaginary parts

etc
etc

but most 4unit students wouldnt have seen that yet, most school leave binomial theorm as the last chapter of 3unit math

#### ella-m

##### New Member
Thanks Heapppss bored of fail 2 & ohexploitable.... your help is much appreciated

#### ella-m

##### New Member
$\bg_white \\\text{34) i. }x+iy+\frac{1}{x+iy}\\=x+iy+\frac{x-iy}{x^2+y^2}\\=\frac{x(x^2+y^2)+x}{x^2+y^2}+\frac{(y(x^2+y^2)-y)i}{x^2+y^2}\\=\frac{x(x^2+y^2+1)}{x^2+y^2}+\frac{y(x^2+y^2-1)i}{x^2+y^2}$
but can you also do 34 part ii?

#### Bored Of Fail 2

##### Banned
but can you also do 34 part ii?
following from that.

now it tells us that this is a real number/expression ( it says z + 1/z = k, where k is real )

so that means its imaginary part must be zero

therefore y ( x^2 +y^2 -1) / ( x^2 + y^2 ) =0

and setting each factor equal to zero gives

y=0 , or x^2 +y^2 =1

im so cluesless on the other part :|, been try to sub in the values and shit and still cnt get anywhere

Last edited:

#### deterministic

##### Member
Suppose y=0, then z+1/z=x+1/x=k
so |k|=|x+1/x|=|(x^2+1)/x|
we use the basic inequality of x^2+1>=2x (Based on (x-1)^2>=0)
Then |k| = |(x^2+1)/x| >= |(2x)/x|=2 so |k|>=2

Suppose x^2+y^2=1:
Then |x|<=1 (as x and y are real numbers)
so subbing (x^2+y^2=1) into the equation for k will give us:
k= x(1+1)/1= 2x
Since |x|<=1, then |k|= |2x|=2|x|<=2*1=2 so |k|<=2

#### ohexploitable

##### hey
Suppose y=0, then z+1/z=x+1/x=k
so |k|=|x+1/x|=|(x^2+1)/x|
we use the basic inequality of x^2+1>=2x (Based on (x-1)^2>=0)
Then |k| = |(x^2+1)/x| >= |(2x)/x|=2 so |k|>=2

Suppose x^2+y^2=1:
Then |x|<=1 (as x and y are real numbers)
so subbing (x^2+y^2=1) into the equation for k will give us:
k= x(1+1)/1= 2x
Since |x|<=1, then |k|= |2x|=2|x|<=2*1=2 so |k|<=2
I gave your post some protection

$\bg_white \\z+\frac1z=\frac{x(x^2+y^2+1)}{x^2+y^2}+\frac{y(x^2+y^2-1)i}{x^2+y^2}\\\text{If it is real then, Im}\left(z+\frac1z\right)=0\\\frac{y(x^2+y^2-1)i}{x^2+y^2}=0\\y(x^2+y^2-1)=0\\x^2+y^2=1\text{ or }y=0$

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$\bg_white \\z+\frac1z=x+iy+\frac{1}{x+iy}$

$\bg_white \\\text{If }y=0,\\z+\frac1z=x+0i+\frac{1}{x+0i}=x+\frac1x$

$\bg_white \\|k|=\left|x+\frac1x\right|\\|k|=\left|\frac{x^2+1}x\right|$

$\bg_white \\(x-1)^2\geq0\\x^2+1\geq2x$

$\bg_white \\|k|=\left|\frac{x^2+1}x\right|\geq\left|\frac{2x}x\right|\\|k|\geq2$

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$\bg_white \\z+\frac1z=\frac{x(x^2+y^2+1)}{x^2+y^2}+\frac{y(x^2+y^2-1)i}{x^2+y^2}$

$\bg_white \\\text{If }x^2+y^2=1,\\z+\frac1z=\frac{x(1+1)}{1}+\frac{y(1-1)i}{1}=2x$

$\bg_white \\x^2+y^2=1,\,|x|\leq1$

$\bg_white \\|k|=2|x|\\\text{since }|x|\leq1\\|k|\leq2$

#### ella-m

##### New Member
I gave your post some protection

$\bg_white \\z+\frac1z=\frac{x(x^2+y^2+1)}{x^2+y^2}+\frac{y(x^2+y^2-1)i}{x^2+y^2}\\\text{If it is real then, Im}\left(z+\frac1z\right)=0\\\frac{y(x^2+y^2-1)i}{x^2+y^2}=0\\y(x^2+y^2-1)=0\\x^2+y^2=1\text{ or }y=0$

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$\bg_white \\z+\frac1z=x+iy+\frac{1}{x+iy}$

$\bg_white \\\text{If }y=0,\\z+\frac1z=x+0i+\frac{1}{x+0i}=x+\frac1x$

$\bg_white \\|k|=\left|x+\frac1x\right|\\|k|=\left|\frac{x^2+1}x\right|$

$\bg_white \\(x-1)^2\geq0\\x^2+1\geq2x$

$\bg_white \\|k|=\left|\frac{x^2+1}x\right|\geq\left|\frac{2x}x\right|\\|k|\geq2$

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$\bg_white \\z+\frac1z=\frac{x(x^2+y^2+1)}{x^2+y^2}+\frac{y(x^2+y^2-1)i}{x^2+y^2}$

$\bg_white \\\text{If }x^2+y^2=1,\\z+\frac1z=\frac{x(1+1)}{1}+\frac{y(1-1)i}{1}=2x$

$\bg_white \\x^2+y^2=1,\,|x|\leq1$

$\bg_white \\|k|=2|x|\\\text{since }|x|\leq1\\|k|\leq2$

Rofls ! thanks for your help... and the protection