Complex Number Question (1 Viewer)

pikachu975 · Nov 20, 2016

If mod (z1+z2) = mod(z1) + mod(z2), show algebraically that arg(z1) = arg(z2). (3 marks)

I was thinking doing like arg(z1) - arg(z2) = 0
arg(z1/z2) = 0 but not sure if it even does anything.

He-Mann · Nov 20, 2016

leehuan · Nov 20, 2016

Ugh. Scratch that I have no idea what I was thinking there with the geometry. Probably cause I only thought of arg(z)=0, pi

seanieg89 · Nov 20, 2016

leehuan said:
Note that arg(z₂) could equal -arg(z₁)

Care to provide an example where this can occur?

seanieg89 · Nov 20, 2016

Mahan1 · Nov 21, 2016

pikachu975 said:
If mod (z1+z2) = mod(z1) + mod(z2), show algebraically that arg(z1) = arg(z2). (3 marks)

I was thinking doing like arg(z1) - arg(z2) = 0
arg(z1/z2) = 0 but not sure if it even does anything.

This is a more trigonometry based solution:

letting $z_{1} = A_{1} e^{i\theta_{1}} , z_{2}= e^{i\theta_{2}}$ and using our assumption we get $z_{1} + z_{2} = (A_{1} + A_{2})e^{i\phi}$

$z_{1} + z_{2} = A_{1} e^{i\theta_{1}} + A_{2} e^{i\theta_{2}} = (A_{1} + A_{2} ) e^{i\phi}$

dividing by $A_{1} e^{i\theta_{1}}$ yields:

$(1+\frac{A_{2}} {A_{1}})e^{i(\phi - \theta_{1})} - \frac{A_{2}}{A_{1}}e^{i(\theta_{2} - \theta_{1})}=1$

equating real and imaginary parts :

$(1+\frac{A_{2}}{A_{1}})cos(\phi - \theta_{1}) - \frac{A_{2}}{A_{1}} cos(\theta_{2} - \theta_{1}) =1$

$(1+\frac{A_{2}}{A_{1}}) sin(\phi - \theta_{1}) - \frac{A_{2}}{A_{1}} sin(\theta_{2} - \theta_{1}) = 0$

squaring real and imaginary parts and using trig identity for $cos(A-B)$ we get

$(1-cos(\phi - \theta_{2}))(2\frac{A_{2}}{A_{1}} + \frac{A^{2}_{2}}{A^{2}_{1}} ) = 0$

but the As are positive so

$(1-cos(\phi - \theta_{2})) = 0$ which means $\phi = \theta_{2}$ with the similar argument we get $\phi = \theta_{1}$

Complex Number Question (1 Viewer)

pikachu975

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He-Mann

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leehuan

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seanieg89

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seanieg89

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Mahan1

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