# Thread: Maximum Modulus of Complex Function

1. ## Maximum Modulus of Complex Function

How do we find the maximum value of $|z^2+2z-3|$ in $|z|=1$?
Can we just simply use the triangle inequality?

2. ## Re: Maximum Modulus of Complex Function

You can use the triangle inequality to get an upper bound for this, but I am unsure whether that is the maximum value though...

3. ## Re: Maximum Modulus of Complex Function

$|z|=1 \Rightarrow z=\cos(\theta) + i \sin(\theta)$

Find a formula for $|z^2 + 2z - 3|^2$ and plug the above $z$ into that.

That gives you a formula in $\theta$ for the square of the modulus.

You can maximize this and then double check that the answers correspond to a maximum for the modulus itself.

Spoiler:
Letting software do all the grubby work

$|z^2 + 2z - 3|^2 = x^4+4 x^3+2 x^2 \left(y^2-1\right)+4 x \left(y^2-3\right)+y^4+10 y^2+9$

with $z = \cos(\theta + i \sin(\theta)$ this becomes (after a ton of computer aided simpilfication)

$|z^2 + 2z - 3|^2 = 8 \sin ^2\left(\frac{\theta }{2}\right) (3 \cos (\theta )+5)$

$\dfrac{d}{d\theta} \left( 8 \sin ^2\left(\frac{\theta }{2}\right) (3 \cos (\theta )+5) \right)=8 \sin \left(\frac{\theta }{2}\right) \left(5 \cos \left(\frac{\theta }{2}\right)+3 \cos \left(\frac{3 \theta }{2}\right)\right)$

setting this equal to 0 and solving for $\theta$ we get

$\theta = 0, ~\pi, ~\pi - \tan^{-1}(2\sqrt{2}), -\pi - \tan^{-1}(2\sqrt{2})$

I leave it to you to determine which of these are maxima and which are local minima using the 2nd derivative.

4. ## Re: Maximum Modulus of Complex Function

Originally Posted by wps
How do we find the maximum value of $|z^2+2z-3|$ in $|z|=1$?
Can we just simply use the triangle inequality?
$|z^2|=1~,~|2z|=2~\&,~|-3|=3$ so $|z^2+2z-3|\le 6$ is true.

Originally Posted by Prove It
You can use the triangle inequality to get an upper bound for this, but I am unsure whether that is the maximum value though...
Good point. If $z=\bf{i}$ then
$|\bf{i}^2+2\bf{i}-3|=|-4+2\bf{i}|=\sqrt{16+4}=2\sqrt5<4.5$

On the other hand, $z=-1$ then $\left|(-1)^2+2(-1)-3\right|=4$

If $z=a+b\bf{i}$ such that |a^2+b^2=1| then $\left|z^2+2z-3\right|^2=\left(a^2-b^2+2a-3\right)^2+\left(2ab+2a\right)^2$.
Can we maximize this?

5. ## Re: Maximum Modulus of Complex Function

Originally Posted by Prove It
You can use the triangle inequality to get an upper bound for this, but I am unsure whether that is the maximum value though...
Originally Posted by Plato
$|z^2|=1~,~|2z|=2~\&,~|-3|=3$ so $|z^2+2z-3|\le 6$ is true.

Good point. If $z=\bf{i}$ then
$|\bf{i}^2+2\bf{i}-3|=|-4+2\bf{i}|=\sqrt{16+4}=2\sqrt5<4.5$

On the other hand, $z=-1$ then $\left|(-1)^2+2(-1)-3\right|=4$

If $z=a+b\bf{i}$ such that |a^2+b^2=1| then $\left|z^2+2z-3\right|^2=\left(a^2-b^2+2a-3\right)^2+\left(2ab+2a\right)^2$ $\rightarrow$ $\mathbf{\left(2ab+2b\right)^2$}.
Can we maximize this?
Yes,got it.
$\left|z^2+2z-3\right|^2=\left( a^2-(1-a^2)+2a-3\right)^2+(1-a^2)\left(2a+2\right)^2$
So we need to maximize a one variable function.
Thanks everyone