1

Ok, so basically I have this:

$$ \frac{3+4i}{5e^{-3i}} $$

So basically, I converted the numerator into polar form and then converted it to exponent form using Euler's formula, but I can have two possible solutions.

I can have $5e^{0.972i}$ (radian) and $5e^{53.13i}$ (angle).

So my real question is, the exponent, does it have to be in radian or angle?

I ended up with $e^{3.927i}$.

colormegone
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Gladstone Asder
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  • Not quite sure what you're asking... the exponent of a complex number in modulus-argument (Euler) form has the angle the complex number makes with the real axis as its imaginary component, measured in radians. If you're asking whether the argument should be stated in radians or degrees, it is conventional and natural to use radians. – Zen May 29 '13 at 19:09
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    One simply does not use degrees for manipulations such as this. – J. M. ain't a mathematician May 29 '13 at 19:10
  • In most cases, one should write out an expression for strange numbers rather than an approximation. $0.972$ and $3.927$ don't really explain anything. – dfeuer May 29 '13 at 19:35
  • Notice that $(3+4i)/5$ is on the unit circle, so that you know already that the final value also is on the unit circle. – Lubin May 29 '13 at 22:05

4 Answers4

3

Put the numerator also in e-power form. r = 5 and the argument follows from arctan(4/3) Now you have an e-power in the Numerator and an e-power in the Denominator. Divide the r-values, subtract the arguments. Don't expect a nice looking answer; arctan(4/3) does not come out good

imranfat
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2

the exponent is always is radians. converting into polar form would give you $$r e^{i \theta}$$ where $$r=\sqrt{3^2 + 4^2}=5$$ and $$\theta=\tan^{-1}(\frac43)$$ or $\arctan (\frac43)$ which gives $$\theta=0.927$$ so your fraction becomes $$\frac{5e^{0.927i}}{5e^{-3i}}$$ $$=e^{3.927i}$$

Ayush Khemka
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2

A complex number $z=x+iy$ can be written as $z=re^{i\theta}$, where $r=|z|=\sqrt{x^2+y^2}$ is the absolute value of $z$ and $\theta=\arg{z}=\operatorname{atan2}(y,x)$ is the angle between the $x$-axis and $z$ measured counterclockwise and in radians. In this case, we have $r=5$ and $\theta=\arctan\frac{4}{3}$ (since $x>0$, see atan2 for more information), so $3+i4=5e^{i\arctan\frac{4}{3}}$ and

$$\frac{3+i4}{5e^{-i3}}=\frac{5e^{i\arctan\frac{4}{3}}}{5e^{-i3}}=e^{i(\arctan\frac{4}{3}+3)}\approx e^{i3.927}.$$

Librecoin
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0

Radians are indeed conventional, and using degrees will lead to trouble once you start messing around with derivatives and integrals. Whenever an angle appears "bare" in a mathematical formula (without, e.g., a degree symbol, indication of gradians, turns, etc.), it is assumed to be in radians.

You can either get both numbers into rectangular form or both into polar form.

With both in polar form, $$\frac{re^{i\alpha}}{qe^{i\beta}} = \frac r q e^{i\alpha} e^{-i\beta} = \frac r q e^{i(\alpha - \beta)}.$$

With both in rectangular form, $$\frac{a+bi}{c+di} = \frac{a+bi}{c+di}\frac{c-di}{c-di} = \frac{(a+bi)(c-di)}{c^2+d^2} = \frac{ac+bd}{c^2+d^2}+\frac{bc-ad}{c^2+d^2}i.$$

dfeuer
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