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complex numbers

how can i do a)? i can convert into degrees but i'm not sure where to go from there.

Edit: or is it easier to draw a diagram after i found the angle and work out from there what all those terms are in the form a+bi? is that easier?

so far(all to 2d.p) i got z1 as 5.80+1.55i
and z2 as -1+root3i

and multiplying those 2 together i get

5.80x-1=-5.80
1.55xroot3=2.69

so the result is -5.80+2.69i to 2.d.p

is that right?

Recall:

\cos(\theta) + i \sin(\theta) = e^{i\theta}

And

r_1e^{i\theta_1} \times r_2e^{i\theta_2} = r_1r_2 \times e^{i(\theta_1+\theta_2)}

Study Forum Helper

Original post by pouii

how can i do a)? i can convert into degrees...

Never... EVER do that with complex numbers!

Forget degrees even exist and you should be good to go.

Do you not know of the formula or general rule? so if you were to multiply lets say z1 and and z2 you would get 12(cos(pi/12 + 2pi/3) + isin(pi/12 + 2pi/3) - so you multiply the non theta values and add the value of theta

for division the theta here for z1/z3 would be pi/12 - 5pi/12 and the factor outside would be 6/3 =2

OP

Original post by Math12345

Recall:

\cos(\theta) + i \sin(\theta) = e^{i\theta}

And

r_1e^{i\theta_1} \times r_2e^{i\theta_2} = r_1r_2 \times e^{i(\theta_1+\theta_2)}

??????? huh it's only fp1 and i don't quite understand.... unless this stuff isn't fp1??

Study Forum Helper

Original post by pouii

??????? huh it's only fp1 and i don't quite understand.... unless this stuff isn't fp1??

Which exam board are you?

Original post by pouii

??????? huh it's only fp1 and i don't quite understand.... unless this stuff isn't fp1??

It's FP2. All my previous post says is that when you multiply z1 and z2 you multiply the value outside the brackets and you add the angles.

(edited 7 years ago)

Remember that

| z_1.z_2|=|z_1||z_2|

and

\text{arg}(z_1.z_2)=\text{arg}(z_1)+\text{arg}(z_2)

.

Study Forum Helper

Original post by pouii

...

You should probably had derived the following statements:

If

z_1=a(cos\theta + isin\theta)

z_2=b(cos\phi + isin\phi)

Then

z_1z_2=ab[cos(\theta + \phi) + isin(\theta + \phi)]

\frac{z_1}{z_2}=\frac{a}{b} [cos(\theta - \phi) + isin(\theta - \phi)]

(edited 7 years ago)

OP

Original post by RDKGames

Which exam board are you?

edexcel

OP

Original post by Math12345

It's FP2. All my previous post says is that when you multiply z1 and z2 you multiply the value outside the brackets and you add the angles.

Original post by B_9710

Remember that

| z_1.z_2|=|z_1||z_2|

and

\text{arg}(z_1.z_2)=\text{arg}(z_1)+\text{arg}(z_2)

.

thanks for the help, picking this question of off google wasn't the best idea....

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