How do you write the complex number in trigonometric form $- 4 + 4 i$ $-4+4i$ ?

Question

3669 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-26T15:47:06

The answer is $= 4 \sqrt{2} (cos (\frac{3 π}{4}) + i sin (\frac{3 π}{4}))$ $=4sqrt2(cos((3pi)/4)+isin((3pi)/4))$

Let $z = a + i b$ $z=a+ib$ be a complex number.

To convert to trigonometric form

$z = r (cos θ + i sin θ)$ $z=r(costheta+isintheta)$

We calculate the modulus, $∥ ∥ z ∥ = \sqrt{a^{2} + b^{2}}$ $∥z∥=sqrt(a^2+b^2)$

$z = - 4 + 4 i$ $z=-4+4i$

$∥ ∥ z ∥ = \sqrt{16 + 16} = \sqrt{32} = 4 \sqrt{2}$ $∥z∥=sqrt(16+16)=sqrt32=4sqrt2$

$z = 4 \sqrt{2} (- \frac{4}{4 \sqrt{2}} + \frac{4 i}{4 \sqrt{2}})$ $z=4sqrt2(-4/(4sqrt2)+(4i)/(4sqrt2))$

$= 4 \sqrt{2} (- \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}})$ $=4sqrt2(-1/sqrt2+i/sqrt2)$

So, $r = 4 \sqrt{2}$ $r=4sqrt2$

$cos θ = - \frac{1}{\sqrt{2}}$ $costheta=-1/sqrt2$

$sin θ = \frac{1}{\sqrt{2}}$ $sintheta=1/sqrt2$

We are in the second quadrant

$θ = \frac{3 π}{4}$ $theta=(3pi)/4$

$z = 4 \sqrt{2} (cos (\frac{3 π}{4}) + i sin (\frac{3 π}{4}))$ $z=4sqrt2(cos((3pi)/4)+isin((3pi)/4))$

How do you write the complex number in trigonometric form -4+4i−4+4i-4+4i?