Solve The Complex Equation

How can I solve this complex equation? [math]z^4+4|z|^2=0[/math]

Let’s start by proving the following lemma:Lemma 1: If [math]z^2[/math] is purely real, then [math]z[/math] is purely real or purely imaginary.Proof: Let [math]z = (a, b) \implies (a^2 - b^2, 2ab)[/math] is real [math]\implies ab = 0 \implies a = 0, or b = 0.[/math]Thus, it follows that in our case, since [math]z^4[/math] must be purely real, [math]z^2[/math] must be either purely real or purely imaginary. Let’s prove another lemma.Lemma 2: If [math]z^2 is[/math] purely imaginary, then it’s real and imaginary parts are equal in magnitude.Proof: Let [math]z = (a, b) \implies (a^2 - b^2, 2ab)[/math]. Since it follows that [math]a^2 - b^2 = 0[/math], then [math]a = \pm b[/math].Thus, we have three cases for [math]z[/math].Case 1: [math]z[/math] is purely real:That is, it is of the form [math](a, 0)[/math]. We have[math]a^2(a^2 + 4) = 0[/math]Since [math]a[/math] is a real number, the only solution is [math]a = 0[/math].Case 2: [math]z[/math] is purely imaginary:That is, it is of the form [math](0, a)[/math]. Then, this also gives us[math]a^4 + 4a^2 = 0[/math],which is identical to Case [math]1[/math].Case 3: [math]z[/math] is of the form [math](a, \pm a)[/math]:We obtain:[math]-4a^4 + 8a^2 = 0[/math]Whence, [math]a = 0[/math] or [math]a = \pm \sqrt{2}.[/math]Thus, the complete set of solutions for [math]z[/math] is:[math]\{(0, 0), (\sqrt{2}, \sqrt{2}), (-\sqrt{2}, \sqrt{2}), (\sqrt{2}, -\sqrt{2}), (-\sqrt{2}, -\sqrt{2})\}[/math]

How do I solve this complex equation step by step?

Here are some tips1) check your answers by substituting your answer back into the equaitons2) a fraction like -4/5 is unlikely in this sort of problem so should hint that you should chck as soon as you found that answer for b3) another way of checking simultaneous equatins is to eliminate another ay, eg here you could eliminate b to determine a. and then find b. This approach may be better than just reading through what you have already written, because in that approach you may just make the same mistake again.Another way of avoiding mistakes is to do things in smaller stapes.in this case you could have toaken the step to 0=8b+8you can do it

How can I solve complex number equation?

Sorry, I didn't notice the first time that you can immediately cancel one z from the first fraction, making the whole thing simply linear in z after multiplying through by the least common multiple of the denominators, which is thus simply z(z+i), (take care to note that this means that if the final answer you get is z=0 or z=-i, these in fact cannot be part of the solution set).  Also, cancel appropriately when performing the "multiply through" (to get rid of the denominators, which is why you're multiplying through in the first place).  Solve the result the way you would a linear equation w/ real coefficients, by "isolating z" using appropriate inverse operations.   HTH (and sorry for the initial misdirection).

How do I solve this complex number equation?

Transform [math]1+i \sqrt 3[/math] into its exponential form ([math]1+i \sqrt{3}=2e^{\arctan \left (\sqrt{3} \right )}=2e^{\frac{i\pi}{3}}[/math]) and than [math]\left (1+i \sqrt 3 \right )^{21}=2^{21} \cdot e^{7 \pi \cdot i}[/math] than you go back to the polar form and take [math]a[/math] from there ([math]a=2^{21} \cdot \cos(7 \pi)=-2 \ 097 \ 152[/math]) so [math]a^2=4 \ 398 \ 046 \ 511 \ 104[/math] and [math]\log_2(a^2)=42[/math]. So the answer is [math]A[/math]. I hope you are allowed a good calculator.I also calculated the full result on Wolfram Alphalooks like [math]b=0[/math].