Given a complex number z:
![z=a+bi](https://img.qammunity.org/2023/formulas/mathematics/college/vx2zgbpz4l566hvwixdkjk6ajx6hz4c4yg.png)
We can write this number in trigonometric form, using:
![\begin{gathered} z=r(\cos\theta+i\sin\theta) \\ . \\ a=r\cos\theta \\ . \\ b=r\sin\theta \\ . \\ r=√(a^2+b^2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4wtnid2bi5698tdohwst41puaaf3wv0l6b.png)
In this case, we are given:
![z=-8+2i](https://img.qammunity.org/2023/formulas/mathematics/college/37el4ff72d4205hoo2xl8u9rvehwlyuxoo.png)
We need to find r and θ. We know:
· a = -8
· b = 2
Thus:
![r=√((-8)^2+2^2)=√(64+4)=√(68)=2√(17)](https://img.qammunity.org/2023/formulas/mathematics/college/5ja4wxa174p7nc8pfxjqtex9w8ptiltnb5.png)
And now, we can find θ:
![-8=2√(17)\cos\theta](https://img.qammunity.org/2023/formulas/mathematics/college/qhppbo6q02mjfasf1iw3kvcy4wo0t2bwhy.png)
And solve:
![\begin{gathered} \cos\theta=-(8)/(2√(17)) \\ . \\ \theta=\cos^(-1)(-(4√(17))/(17))\approx2.89661 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/cc6u4yuh0wfl7dub23dfcikkzr26jbax18.png)
Now we can write the number in trigonometric form:
![z=2√(17)(\cos(2.9)+i\sin(2.9))](https://img.qammunity.org/2023/formulas/mathematics/college/cul1myn4dxrszel9gy1h9ve3ynehu2l45w.png)