danbane
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Really stuck on this complex numbers question during my revision, goes as follows:

z1 = 2+3i
l z1z2 l = 39root2 (l z1z2 l refers to modulus of z1 * z2)
arg(z1z2) = pi/4

b) find z2, giving your answer in the form a+ib where a and b are integers (6 marks)

Any help would be greatly appreciated !
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DFranklin
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I think the easiest thing here is to find z_1z_2 in x+iy form, and then from there divide by z1 to find z2.
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RDKGames
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(Original post by danbane)
Really stuck on this complex numbers question during my revision, goes as follows:

z1 = 2+3i
l z1z2 l = 39root2 (l z1z2 l refers to modulus of z1 * z2)
arg(z1z2) = pi/4

b) find z2, giving your answer in the form a+ib where a and b are integers (6 marks)

Any help would be greatly appreciated !
From the info you should begin by finding the modulus and argument of z2.
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cactus11235813
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(Original post by danbane)
Really stuck on this complex numbers question during my revision, goes as follows:

z1 = 2+3i
l z1z2 l = 39root2 (l z1z2 l refers to modulus of z1 * z2)
arg(z1z2) = pi/4

b) find z2, giving your answer in the form a+ib where a and b are integers (6 marks)

Any help would be greatly appreciated !
There are a bunch of rules about this that can help, like:
|z1 z2| = |z1| |z2|
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danbane
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(Original post by username11235813)
There are a bunch of rules about this that can help, like:
|z1 z2| = |z1| |z2|
so i tried this rule and divided 39root2 by lz1l and got 3root26 which should be the modulus of z2 but how could i find the a+ib format of z2 from here?

thanks for the tip btw
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danbane
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(Original post by DFranklin)
I think the easiest thing here is to find z_1z_2 in x+iy form, and then from there divide by z1 to find z2.
do you know how you would go about finding z1z2 in x+iy form?
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DFranklin
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(Original post by danbane)
do you know how you would go about finding z1z2 in x+iy form?
You know the modulus and argument of z1z2. In particular, you should be able to *write down* a complex number of argument pi/2. You then just need to scale it to make the modulus match the given modulus for z1z2.
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danbane
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(Original post by DFranklin)
You know the modulus and argument of z1z2. In particular, you should be able to *write down* a complex number of argument pi/2. You then just need to scale it to make the modulus match the given modulus for z1z2.
Ok so found both the modulus and argument of z2, arranged into mod-arg form, inputted the mod-arg form into my calculator and got nice integers. My final answer for a+ib was 15-3i, could you check if I'm correct, thank you in advance!
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davros
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(Original post by danbane)
Ok so found both the modulus and argument of z2, arranged into mod-arg form, inputted the mod-arg form into my calculator and got nice integers. My final answer for a+ib was 15-3i, could you check if I'm correct, thank you in advance!
I get 15 - 3i (or 3(5 - i)) as the answer too

Did you write z_1z_2 = re^{i\pi/4} where r = |z_1z_2| and then divide by z_1 to find z_2? The square root of 2 cancels out nicely when you do this
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danbane
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(Original post by davros)
I get 15 - 3i (or 3(5 - i)) as the answer too

Did you write z_1z_2 = re^{i\pi/4} where r = |z_1z_2| and then divide by z_1 to find z_2? The square root of 2 cancels out nicely when you do this
no, but that certainly sounds like an interesting method. Thanks for checking my answer!
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DFranklin
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(Original post by davros)
Did you write z_1z_2 = re^{i\pi/4} where r = |z_1z_2| and then divide by z_1 to find z_2? The square root of 2 cancels out nicely when you do this
I'm not sure how much I'm being "an old fuddy duddy", but I'm somewhat despairing at the "convert everything into mod/arg form and let the calculator sort it out" mentality on display by some in this thread...
Last edited by DFranklin; 3 weeks ago
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danbane
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(Original post by DFranklin)
I'm not sure how much I'm being "an old fuddy duddy", but I'm somewhat despairing at the "convert everything into mod/arg form and let the calculator sort it out" mentality on display by some in this thread...
In fairness, I was a bit uneasy with this method considering the question is worth 6 marks. Do you think I would be able to achieve the full 6 marks using this?
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davros
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(Original post by danbane)
In fairness, I was a bit uneasy with this method considering the question is worth 6 marks. Do you think I would be able to achieve the full 6 marks using this?
I personally am not sure, but the calculation comes out very easily when done ,manually and I also would not expect a calculator to be used for this!
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