What is the trigonometric form of $(2 - i) \cdot (1 + i)$ $(2-i)*(1+i)$ ?

Question

2184 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-15T13:21:19

See explanation.

First we have to multiply the numbers:

$(2 - i) \cdot (1 + i) = 2 + 2 i - i - i^{2} = 2 + i + 1 = 3 + i$ $(2-i)*(1+i)=2+2i-i-i^2=2+i+1=3+i$

Now we have to write the result $3 + i$ $3+i$ in trigonometric form:

$| z | = \sqrt{3^{2} + 1^{2}} = \sqrt{10}$ $|z|=sqrt(3^2+1^2)=sqrt(10)$

Now we have to find the angle using:

$cos φ = \frac{r e (z)}{| z |}$ $cosvarphi=(re(z))/|z|$

$cos φ = \frac{3}{\sqrt{10}} = \frac{3 \sqrt{10}}{10} \approx 0.949$ $cosvarphi=3/sqrt(10)=(3sqrt(10))/10~~0.949$

From this we have $φ \approx {18.43}^{o}$ $varphi ~~18.43^o$ .

Finally we can write the result as:

$z = \sqrt{10} \cdot ({cos 18.43}^{o} + i {sin 18.43}^{o})$ $z=sqrt(10)*(cos18.43^o + isin18.43^o)$

What is the trigonometric form of (2−i)⋅(1+i) (2-i)*(1+i) ?