What is the trigonometric form of $12 + 4 i$ $12+4i$ ?

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What is the trigonometric form of $12 + 4 i$ $12+4i$ ?

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Answer 1 · 2016-05-19T09:50:17

$12 + 4 i = 4 \sqrt{10} (cos α + i sin α)$ $12+4i=4sqrt10(cosalpha+isinalpha)$ , where $tan α = \frac{1}{3}$ $tanalpha=1/3$

Explanation:

$a + i b$ $a+ib$ can be written in trigonometric form $r e^{i θ} = r cos θ + i r sin θ = r (cos θ + i sin θ)$ $re^(itheta)=rcostheta+irsintheta=r(costheta+isintheta)$ ,
where $r = \sqrt{a^{2} + b^{2}}$ $r=sqrt(a^2+b^2)$ .

Hence $12 + 4 i = \sqrt{12^{2} + 4^{2}} [cos α + i sin α]$ $12+4i=sqrt(12^2+4^2)[cosalpha+isinalpha]$

= $\sqrt{160} [cos α + i sin α]$ $sqrt160[cosalpha+isinalpha]$

= $4 \sqrt{10} [cos α + i sin α]$ $4sqrt10[cosalpha+isinalpha]$ ,

where $cos α = \frac{12}{4 \sqrt{10}} = \frac{3}{\sqrt{10}}$ $cosalpha=12/(4sqrt10)=3/sqrt10$

and $sin α = \frac{4}{4 \sqrt{10}} = \frac{1}{\sqrt{10}}$ $sinalpha=4/(4sqrt10)=1/sqrt10$

or $tan α = \frac{\frac{1}{\sqrt{10}}}{\frac{3}{\sqrt{10}}} = \frac{1}{3}$ $tanalpha=(1/sqrt10)/(3/sqrt10)=1/3$

Answer 2 · 2016-05-19T10:01:49

$4 \sqrt{10} (cos (0.322) + i sin (0.322))$ $4sqrt10(cos(0.322)+isin(0.322))$

Explanation:

Given a complex number z = x + yi , this can be written in trig. form as.

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

where $r = \sqrt{x^{2} + y^{2}}$ $r=sqrt(x^2+y^2)$

and $θ = {tan}^{- 1} (\frac{y}{x})$ $theta=tan^-1(y/x)$

here x = 12 and y = 4

$\Rightarrow r = \sqrt{12^{2} + 4^{2}} = \sqrt{160} = 4 \sqrt{10}$ $rArrr=sqrt(12^2+4^2)=sqrt160=4sqrt10$

and $θ = {tan}^{- 1} (\frac{4}{12}) \approx 0.322 radians or {18.43}^{\circ}$ $theta=tan^-1(4/12)≈0.322" radians or"18.43^@$

$\Rightarrow 12 + 4 i = 4 \sqrt{10} (cos (0.322) + i sin (0.322))$ $rArr12+4i=4sqrt10(cos(0.322)+isin(0.322))$ or

$12 + 4 i = 4 \sqrt{10} ({cos (18.43)}^{\circ} + i {sin (18.43)}^{\circ})$ $12+4i=4sqrt10(cos(18.43)^@+isin(18.43)^@)$

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What is the trigonometric form of $12 + 4 i$ $12+4i$ ?

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What is the trigonometric form of 12+4i 12+4i ?

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What is the trigonometric form of $12 + 4 i$ $12+4i$ ?