Complex q (1 Viewer)

axyx · Nov 5, 2023

Prove true for n = 3
$z_3=(1+i)\left(1+\frac{i}{|1+i|}\right)$
$=(1+i)\left(1+\frac{i}{\sqrt{2}}\right)$
$=\frac{\sqrt{2}-1}{\sqrt{2}}+\left(\frac{1+\sqrt{2}}{\sqrt{2}}\right)i$
$|z_3|=\sqrt{\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)^2+\left(\frac{1+\sqrt{2}}{\sqrt{2}}\right)^2}$
$=\sqrt{3}$

Assume true for n = k
$|z_k|=\sqrt{k}$

Prove true for n = k+1
$LHS=|z_{k+1}|$
$RHS=\sqrt{k+1}$
$z_{k+1}=z_k\left(1+\frac{i}{|z_k|}\right)$
Using assumption
$=z_k\left(1+\frac{i}{\sqrt{k}}\right)$
Let $z_k=x+iy$ then
$|z_k|=\sqrt{x^2+y^2}$
$z_{k+1}=(x+iy)\left(1+\frac{i}{\sqrt{k}}\right)$
$=x+\frac{xi}{\sqrt{k}}+iy-\frac{y}{\sqrt{k}}$
$LHS=\sqrt{\left(\frac{x\sqrt{k}-y}{\sqrt{k}}\right)^2+\left(\frac{y\sqrt{k}+x}{\sqrt{k}}\right)^2}$
$=\frac{1}{\sqrt{k}}\sqrt{x^2k-2xy\sqrt{k}+y^2+y^2k+2xy\sqrt{k}+x^2}$
$=\frac{1}{\sqrt{k}}\sqrt{x^2k+y^2+y^2k+x^2}$
$=\frac{1}{\sqrt{k}}\sqrt{(x^2+y^2)(k+1)}$
$=\frac{1}{\sqrt{k}}|z_k|\sqrt{k+1}$
Using assumption
$=\frac{\sqrt{k}}{\sqrt{k}}\sqrt{k+1}$
$=\sqrt{k+1}$
$=RHS$ as required

Conclusion: Thus by PMI bla blabla $\forall n \in \mathbb{Z}\ge2$

kendricklamarlover101 · Nov 6, 2023

axyx said:
Prove true for n = 3
$z_3=(1+i)\left(1+\frac{i}{|1+i|}\right)$
$=(1+i)\left(1+\frac{i}{\sqrt{2}}\right)$
$=\frac{\sqrt{2}-1}{\sqrt{2}}+\left(\frac{1+\sqrt{2}}{\sqrt{2}}\right)i$
$|z_3|=\sqrt{\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)^2+\left(\frac{1+\sqrt{2}}{\sqrt{2}}\right)^2}$
$=\sqrt{3}$

Assume true for n = k
$|z_k|=\sqrt{k}$

Prove true for n = k+1
$LHS=|z_{k+1}|$
$RHS=\sqrt{k+1}$
$z_{k+1}=z_k\left(1+\frac{i}{|z_k|}\right)$
Using assumption
$=z_k\left(1+\frac{i}{\sqrt{k}}\right)$
Let $z_k=x+iy$ then
$|z_k|=\sqrt{x^2+y^2}$
$z_{k+1}=(x+iy)\left(1+\frac{i}{\sqrt{k}}\right)$
$=x+\frac{xi}{\sqrt{k}}+iy-\frac{y}{\sqrt{k}}$
$LHS=\sqrt{\left(\frac{x\sqrt{k}-y}{\sqrt{k}}\right)^2+\left(\frac{y\sqrt{k}+x}{\sqrt{k}}\right)^2}$
$=\frac{1}{\sqrt{k}}\sqrt{x^2k-2xy\sqrt{k}+y^2+y^2k+2xy\sqrt{k}+x^2}$
$=\frac{1}{\sqrt{k}}\sqrt{x^2k+y^2+y^2k+x^2}$
$=\frac{1}{\sqrt{k}}\sqrt{(x^2+y^2)(k+1)}$
$=\frac{1}{\sqrt{k}}|z_k|\sqrt{k+1}$
Using assumption
$=\frac{\sqrt{k}}{\sqrt{k}}\sqrt{k+1}$
$=\sqrt{k+1}$
$=RHS$ as required

Conclusion: Thus by PMI bla blabla $\forall n \in \mathbb{Z}\ge2$

instead of letting $z_{k}=x+iy$ u could just leave it as $z_{k}$
$LHS=|z_{k+1}|$
$=\left|z_{k}\left(1+\frac{i}{|z_{k}|}\right)\right|$
$=\left|z_{k}\left(\frac{|z_{k}|+i}{|z_{k}|}\right)\right|$
$=||z_{k}|+i|$
Using assumption
$=|\sqrt{k}+i|$
Since $\sqrt{k}$ is real and i is imaginary,
$|\sqrt{k}+i|=\sqrt{(k)^{2}+1}$
$=\sqrt{k^2+1}$
$=RHS$ as required
hence if true for n=k true for n=k+1 blah blah blah yk whats up

Luukas.2 · Nov 6, 2023

kendricklamarlover101 said:
instead of letting $z_{k}=x+iy$ u could just leave it as $z_{k}$
$LHS=|z_{k+1}|$
$=\left|z_{k}\left(1+\frac{i}{|z_{k}|}\right)\right|$
$=\left|z_{k}\left(\frac{|z_{k}|+i}{|z_{k}|}\right)\right|$
$=||z_{k}|+i|$
Using assumption
$=|\sqrt{k}+i|$
Since $\sqrt{k}$ is real and i is imaginary,
$|\sqrt{k}+i|=\sqrt{(k)^{2}+1}$
$=\sqrt{k^2+1}$
$=RHS$ as required
hence if true for n=k true for n=k+1 blah blah blah yk whats up

There's a small error in the end of this. It should say:

Since $\sqrt{k}$ is real and $i$ is imaginary, $\left|\sqrt{k}+i\right|=\sqrt{\left(\sqrt{k}\right)^{2}+1^2}=\sqrt{k+1}=\text{RHS}$ as required

kendricklamarlover101 · Nov 6, 2023

Luukas.2 said:
There's a small error in the end of this. It should say:

Since $\sqrt{k}$ is real and $i$ is imaginary, $\left|\sqrt{k}+i\right|=\sqrt{\left(\sqrt{k}\right)^{2}+1^2}=\sqrt{k+1}=\text{RHS}$ as required

oops yea didnt even notice was too busy trying to figure out latex lol

Luukas.2 · Nov 7, 2023

This recursive definition yields a spiral of points in the Argand diagram... here is $z_n$ for $n \in \{1,\,2,\,3,\, ...,\,37\}$ starting from $z_1 = 1$ .

year10studentpreparin · Nov 7, 2023

i reckon the answer is c)

appleali · Nov 10, 2023

year10studentpreparin said:
i reckon the answer is c)

C is always right

Complex q (1 Viewer)

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