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A complex number z is given by z=(1+2i)^2 +((3+i)/(1+i))
Express z in the form x+iy?
Ive expanded the (1+2i)^2 to get 4i-3 , when i times the other function by the conjugate to have to times it by 4i +3 too?
(edited 7 years ago)

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No.
No
Reply 3
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Lychee628
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Okay thanks guys , ill work it out & ill tell you what i got in a minute
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I got 3i+5 , is that right? Its a 2001 paper , so couldnt fine the mark scheme
Original post by Ayaz789
I got 3i+5 , is that right? Its a 2001 paper , so couldnt fine the mark scheme


No you should get −1+3i-1+3i.You should've got (1+2i)2=−3+4i(1+2i)^2=-3+4i, then you multiply top and bottom of (3+i)/(1+i)(3+i)/(1+i) to get 2−i2-i and add those together.
(edited 7 years ago)
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Original post by IrrationalRoot
No you should get −1+3i-1+3i.You should've got (1+2i)2=−3+4i(1+2i)^2=-3+4i, then you multiply top and bottom of
Unparseable latex formula:

\frac{3+i){1+i}

to get 2−i2-i and add those together.


Its not showing it properly , the message you wrote:/
Original post by Ayaz789
Its not showing it properly , the message you wrote:/


Fixed, Latex is being stupid for some reason.
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Original post by IrrationalRoot
No you should get −1+3i-1+3i.You should've got (1+2i)2=−3+4i(1+2i)^2=-3+4i, then you multiply top and bottom of (3+i)/(1+i)(3+i)/(1+i) to get 2−i2-i and add those together.


Ahh i made a silly mistake at the end , instead of taking away from 2 i added 3 to 2 to get 5:/
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Original post by IrrationalRoot
Fixed, Latex is being stupid for some reason.


Top man!
Reply 10
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Original post by IrrationalRoot
Fixed, Latex is being stupid for some reason.


So is the modulus Root 10?
And the argument is tan-1(3/-1) which is -1.25
Do i do Pi minus that answer to get 1.89?
Original post by Ayaz789
So is the modulus Root 10?
And the argument is tan-1(3/-1) which is -1.25
Do i do Pi minus that answer to get 1.89?


Correct modulus.
π−arctan⁡3\pi-\arctan3 for argument.
Reply 12
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Original post by IrrationalRoot
Correct modulus.
π−arctan⁡3\pi-\arctan3 for argument.


Whats arctan3?
Original post by Ayaz789
Whats arctan3?


Same as tan⁡−13\tan^{-1} 3 i.e. the angle the position vector of the complex number makes with the negative real axis, so you do π\pi minus that to get the argument which is the angle it makes with the positive real axis.
(edited 7 years ago)
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Original post by IrrationalRoot
Same as tan⁡−13\tan^{-1} 3 i.e. the angle the position vector of the complex number makes with the negative real axis, so you do π\pi minus that to get the argument which is the angle it makes with the positive real axis. tan⁡−13\tan^{-1} 3


Is my argument right of 1.89?
Original post by Ayaz789
Is my argument right of 1.89?


Yes, but I hope you understand the method and how it works.
Reply 16
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Original post by IrrationalRoot
Yes, but I hope you understand the method and how it works.


Can you repeat why its not tan (3/-1)
Original post by Ayaz789
Can you repeat why its not tan (3/-1)


It's not tan⁡−1(−3)\tan^{-1}(-3) because that will give you a fourth quadrant angle, and you know the complex number −1+3i-1+3i is in the second quadrant. You would need to add pipi to tan⁡−1(−3)\tan^{-1}(-3) to get the argument.

Alternatively, do what I do and draw a little sketch of the right triangle on an Argand diagram.
Reply 18
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Original post by IrrationalRoot
It's not tan⁡−1(−3)\tan^{-1}(-3) because that will give you a fourth quadrant angle, and you know the complex number −1+3i-1+3i is in the second quadrant. You would need to add pipi to tan⁡−1(−3)\tan^{-1}(-3) to get the argument.

Alternatively, do what I do and draw a little sketch of the right triangle on an Argand diagram.


Ahh okay just the -1 is confusing me thats all!:/
Reply 19
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Zacken
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Original post by Ayaz789
Ahh okay just the -1 is confusing me thats all!:/




If you draw this sort of diagram it becomes clear. The red angle you find using trigonometry, and then the green angle is your argument which you find by doing pi-red angle.

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