# How do you write the complex number z = 6(cos 45° + i sin 45°)  in standard form?

I found: $3 \sqrt{2} + 3 \sqrt{2} i$
We need to evaluate the values of $\sin$ and $\cos$ and multiply by $6$:
$z = 6 \left(\cos \left({45}^{\circ}\right) + i \sin \left({45}^{\circ}\right)\right) = 6 \left(\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}\right) =$
$= {\cancel{6}}^{3} \left(\frac{\sqrt{2}}{\cancel{2}} + i \frac{\sqrt{2}}{\cancel{2}}\right) = 3 \sqrt{2} + 3 \sqrt{2} i$
which is in the form: $a + b i$