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In various places in physics, EM for example, complex numbers are used to describe things that are physically real. I will point a simple case - solving an ODE for resistance/charge/voltage. We get a certain value, which is in the form $a+bi$ and take its real part. My question is, what does the imaginary part of the answer physically represents? How do we know we are not "losing information" by considering only the real part when using complex numbers in calculations? The whole treatment of complex numbers in physics baffles me. What does the complex plane have to do with reality? Sometimes certain motions are also described with complex numbers.

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  • $\begingroup$ "Imaginary" is an unfortunate (historical) choice of naming the part of an ordered-pair number. Also unfortunate is that no one of mathmatical influence is willing to change that name. I, personally, think "transverse" would be better. (Real) +i(Transverse) would be reminiscent of the Argand plane. $\endgroup$ – Bill N Jun 21 '20 at 16:16
  • $\begingroup$ To preserve the symbols $\Bbb{C}$ and $\Bbb{R}$, I suggest the names "Complete numbers" and "Restricted numbers". $\endgroup$ – badjohn Jun 21 '20 at 22:18
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The question is based on a false premise: that only the real part of complex current or voltage is significant.

In fact, it is the magnitude of the complex quantity that is significant. The relationship between the real and imaginary parts tells you the timing or phase of the quantity relative to a pure resistance

If you had a circuit with a resistor, capacitor and inductance arranged in series and fed with $120$ V alternating current, you could calculate that the voltage across the inductance and the capacitor were both purely imaginary. Yet a AC voltmeter could measure a very real voltage across each of these. More importantly, you could get blown on your backside by this imaginary voltage, if you were to touch the capacitor or inductance at the wrong moment in the alternating voltage.

As a different example: If you were to calculate the voltage individually across four circuit elements in series and find $$V_1=(5+0i) \text{ Volts}$$$$V_2=(0+5i) \text{ Volts}$$$$V_3=(0-5i) \text{ Volts}$$$$V_4=(3+4i) \text{ Volts}$$then you would know that the measured voltage across each of the elements would be exactly the same, $5$ Volts.

If you were to display the voltages on a multiple trace oscilloscope, you would see four identical sine waves with an amplitude of $5$ Volts: $V_2$ would be $90$ degrees out of phase with $V_1$, $V_3$ would be $180$ degrees out of phase with $V_2$, and $V_4$ would be $53.13$ degrees out of phase with $V_1$ $( \text{ because }\tan(53.13)=\frac{4}{3})$

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  • $\begingroup$ I agree that in some theories the imaginary part has a well defined „real world“ meaning. But I think in Quantum Mechanics the situation is quite different. Although the mathematics of QM works perfectly, the imaginary part has no meaningful physical imprint (for lack of a better word) in the measurable world. I always found this to be the most mysterious aspect of QM. $\endgroup$ – Hartmut Braun Jun 21 '20 at 16:19
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In the simplest case we may use an exponential like an $A\cdot \text{e}^{\text{i}\omega t}$ instead of a $A\cdot\sin(\omega t)$, so it is just the easiest way of writing the solutions (simplest harmonic oscillations).

But in reality the amplitude and the phase may vary with time in the transient processes, so you must write two equations instead of one. Often these two equations are equivalent to one equation with a complex coefficients and solutions simply connected to the variables in question.

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  • $\begingroup$ Shouldn't we say that the real part and the "imaginary" part are actually used in the form $\mathrm{A} \cdot e^{\iota \omega t}$? Since, any complex number (thanks to Euler) $z=x+\iota y$ can be written in the polar form $\sqrt{x^2+y^2} \cdot e^{\iota \theta}$; where, $\theta= \tan^{-1}(\frac{y}{x})$ (phase). So, we actually have all the information for the solution to an S.H.O differential equation. $\endgroup$ – Ishika_96_sparkle Jun 21 '20 at 16:09
  • $\begingroup$ @Ishika_96_sparkle: Yes, foe a SHO there are two initial conditions and two constants to determine. Linearity and homogenity of SHO makes the exponential form of solution appealing. But in (very) general, it is not always fruitful to join two independent variables in one complex variable $z$. $\endgroup$ – Vladimir Kalitvianski Jun 21 '20 at 16:40
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Certainly you lose information by only taking the real part of a complex number, but sometimes that is the only part you are interested in. In the case of resistance, the complex quantity is called impedance, and the imaginary part (called reactance) covers the effect of capacitors and inductances.

In general complex numbers are used because they make the Maths easier. An unexpected complex answer, like getting $sin\theta>1$ in Snell's Law, tells you something different is happening.

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another example:

to solution of this differential equation:

$${\frac {d^{2}}{d{t}^{2}}}x \left( t \right) +{\omega}^{2}x \left( t \right) +2\,\gamma\,{\frac {d}{dt}}x \left( t \right)=0 $$

is: (Ansatz)

$$x(t)= \left( a+ib \right) {{\rm e}^{ \left( -\gamma+i\sqrt {-{\gamma}^{2}+{ \omega}^{2}} \right) t}}+ \left( a-ib \right) {{\rm e}^{ \left( - \gamma-i\sqrt {-{\gamma}^{2}+{\omega}^{2}} \right) t}} $$

where the imaginary part of x(t) is zero .

Thus : because $\Im(x(t))=0$ , the solution is $\Re(x(t))$

with the initial condition

$$x(0)=x_0\quad ,\dot{x}(0)=v_0$$

you have two equations for $a$ and $b$

$$a=\frac{x_0}{2}$$ $$b=\frac{1}{2}\,{\frac {x_{{0}}\gamma+v_{{0}}}{\sqrt {-{\gamma}^{2}+{\omega}^{2}}}}$$

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