Question

solve z=x+iy when sinz=11

Answer 1

You can try to use the formula:\[\sin z=\frac{ e^{iz}-e^{-iz} }{ 2i }\]So:\[\frac{ e^{iz}-e^{-iz} }{ 2i }=11\]Multiply with 2i and then with \(e^{iz}\) to get a quadratic equation.

Answer 2

thank you but it says "type your answers in terms of pi and n. type your answers in radians and to 4dp"

Answer 3

Well, there is still work to do before we get to that:
\[e^{iz}-e^{-iz}=22i\]multiply with \(e^{iz}\) and tidy up:\[(e^{iz})^2-22i \cdot e^{iz}-1=0\]
Use quadratic formula to find solutions:\[e^{iz}=\frac{ 22i \pm \sqrt{-484+4} }{ 2 }=11i \pm 2 \sqrt{30}i=(11 \pm 2\sqrt{30})i\]

Answer 4

Now you can solve for z by taking the complex logarithm and dividing by i.

Answer 5

thank you but i still dont get it what is the answer?

Answer 6

OK. Definition of complex logarithm: \(\ln z= \ln|z| + i Arg(z)\).

Now \(|(11 \pm 2\sqrt{30})i|=11 \pm2\sqrt{30}\) (both numbers are on the positive imaginary axis).

Then: Arg(\((11 \pm 2\sqrt{30})i)=\frac{\pi}{2}\).
Taking the complex logarithm:\[iz=\ln(11 \pm 2\sqrt{30})+i\frac{\pi}{2}\]
Now divide by i:\[z=\frac{\pi}{2}+\frac{\ln(11 \pm 2\sqrt{30})}{i}=\frac{\pi}{2}-i \ln(11 \pm 2\sqrt{30})\]

Answer 7

We can use a calculator to find decimal values:

\(\ln(11+2\sqrt{30})\approx 3.0890\),
\(\ln(11-2\sqrt{30})\approx -3.0890\),

\(\frac{\pi}{2}\approx 1.5708\)

So: \(z \approx 1.5708 \pm 3.0890i\)

Answer 8

thank you but how is the answer in terms of pi and n?

Answer 9

We had 1/2 pi as a real part.
I'm afraid I don't know what you mean with n.

Answer 10

im not sure either but thank you

Answer 11

BTW, it could mean, because sin z is periodic, that the Arg(z) can also be added with n times 2 pi .

Answer 12

so how would that be included in the answer?

Answer 13

Hold on, I am looking in to that...

Answer 14

alright

Answer 15

It means that \(\frac{\pi}{2}\) must be replaced with \(\frac{\pi}{2}+2n\pi=\frac{(4n+1)\pi}{2}\), so

\(z \approx (4n+1)\cdot 1.5708\pm3.0890i\)

Hope this helps!

Answer 16

Checked results on WolframAlpha - worked out fine :)
See image.

Answer 17

thank you