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[samuelclarke]

if z1 and z2 are any two complex numbers such that lz1 - z2l = lz1 + z2l show that arg(z1) - arg(z2) = pi/2


Another question,

y=x^x

lny=xlnx

show that

d(x^x)/dx = x^x (1 + lnx)

 

[Aesytic]

1. the complex numbers z1 + z2 and z1 - z2 are the diagonals of the parallelogram that's formed when you add z1 and z2
if the moduli of these complex numbers are equal, then the diagonals of the parallelogram are equal, which makes it a rectangle
if it's a rectangle, then the angle between z1 and z2 must be 90 degrees since the angles of a rectangle are right angles
.'. arg(z1) - arg(z2) = pi/2
it becomes a lot clearer if you draw it out on the argand plane

2. lny = xlnx
differentiating implicitly,
(dy/dx)*(1/y) = lnx + 1
dy/dx = y(lnx + 1)
but y = x^x
.'. dy/dx = d(x^x)/dx = x^x (lnx + 1)

 

[Nooblet94]

if z1 and z2 are any two complex numbers such that lz1 - z2l = lz1 + z2l show that arg(z1) - arg(z2) = pi/2


Another question,

y=x^x

lny=xlnx

show that

d(x^x)/dx = x^x (1 + lnx)

Ahh I love that question. First time I saw it was when we did differentiation in year 11 (although it was way beyond anything we were required to do). Stumbled across it right before I went to bed and then managed to solve it whilst lying in bed trying to get to sleep. I was so proud when I looked up the answer the next morning and saw I was right. Anyway;

<a href="http://www.codecogs.com/eqnedit.php?latex=~\\ y=x^x\\ ~\\ $Taking the natural log of both sides and applying the power rule, we get$\\ ~\\ \ln y=x\ln x\\ ~\\ $Differentiating implicitly with respect to $x\\ ~\\ \frac{1}{y}\cdot \frac{dy}{dx}=\frac{x}{x}@plus;\ln x\\ \frac{dy}{dx}=y(1@plus;\ln x)\\ ~\\ $Substituting in from the original equation$\\ ~\\ \frac{dy}{dx}=x^x(1@plus;\ln x)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?~\\ y=x^x\\ ~\\ $Taking the natural log of both sides and applying the power rule, we get$\\ ~\\ \ln y=x\ln x\\ ~\\ $Differentiating implicitly with respect to $x\\ ~\\ \frac{1}{y}\cdot \frac{dy}{dx}=\frac{x}{x}+\ln x\\ \frac{dy}{dx}=y(1+\ln x)\\ ~\\ $Substituting in from the original equation$\\ ~\\ \frac{dy}{dx}=x^x(1+\ln x)" title="~\\ y=x^x\\ ~\\ $Taking the natural log of both sides and applying the power rule, we get$\\ ~\\ \ln y=x\ln x\\ ~\\ $Differentiating implicitly with respect to $x\\ ~\\ \frac{1}{y}\cdot \frac{dy}{dx}=\frac{x}{x}+\ln x\\ \frac{dy}{dx}=y(1+\ln x)\\ ~\\ $Substituting in from the original equation$\\ ~\\ \frac{dy}{dx}=x^x(1+\ln x)" /></a>

 

[samuelclarke]

thanks guys!