How to find $c$ and $d$ from the equation $(c+id)^2=1$?

Question

I need to solve this complex equation: $$ (c + id)^2 = 1 $$

where $i^2=-1$.

What am I supposed to calculate here? Just $c$ and $d$?

Answer 1

$$(c+id)^2=1 \Rightarrow c^2+2cdi+i^2d^2=1 \Rightarrow c^2+2cdi-d^2=1 \Rightarrow (c^2-d^2)+2cdi=1+0i$$

The real part of the LHS must be equal to the real part of the RHS and the imaginary part of the LHS must equal to the imaginary part of the RHS.

So, you have to solve the following system : $$c^2-d^2=1 \\ 2cd=0$$

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From the first equation we get $c^2=1+d^2$.

Substituting this in $2cd=0 \Rightarrow cd=0 \Rightarrow c^2d^2=0$ we get $$(1+d^2)d^2=0 \\ \Rightarrow 1+d^2=0 \text{ or } d^2=0 \\ \Rightarrow d^2=-1 \ \ \text{ or } \ \ d=0 \\ \Rightarrow d=\pm i \ \ \text{ or } \ \ d=0$$

For $d=\pm i$, substituting at $c^2=1+d^2$ we get $c^2=0 \Rightarrow c=0$.

For $d=0$, substistuting at we get $c^2=1 \Rightarrow c=\pm 1$.

Answer 2

Your post says $i^2=1$, surely you mean $i^2=-1$ the complex unit, no?

Well assuming $i^2=-1$, firstly $$(c+id)^{2}=(c+id)(c+id)$$

After multiplying we find $$c^{2}+2icd+(id)^{2}=1$$ Since $i^{2}=-1$ $$c^{2}-d^{2}+2icd=1$$ Since $1$ is real and there is no imaginary part on the r.h.s we can split to two equations \begin{align} c^{2}-d^{2} &= 1 \\ cd &= 0 \end{align} Two equations in two unknowns, can you take it form here?

Best, Bacon.

EDIT In response to your comment below (I don't have sufficient reputation to make a comment yet) I would say that in order for the product of two quantities to be identically zero, at least one needs to be zero. Therefore using this fact and the difference of two squares observation above it, you have several cases to consider before you find the values of $c$ and $d$.

Answer 3

First of all i would observe $$1 = e^{i2k\pi} \;\; k \in \mathbb{Z}$$ then i would write $(a + ib)^2 = \left( \rho e^{i\theta} \right)^2 = \rho^2 e^{i2\theta}$ so...

$$\rho^2 e^{i2\theta} = e^{i2k\pi}$$ which of course implies that $\rho = 1$, (since it is a positive real number) while

$$ 2\theta = 2 k \pi \Rightarrow \theta = k \pi$$

So the solutions are of the form

$$a + ib = e^{ik\pi}$$ which implies for $k$ even $$a + ib = 1 \Rightarrow \left\{ \begin{array}{l} a = 1 \\ b = 0 \end{array} \right.$$

while for $k$ odd

$$a + ib = -1 \Rightarrow \left\{ \begin{array}{l} a = -1 \\ b = 0 \end{array} \right.$$

Answer 4

By the binomial formulas we know that $$ 0=z^2-1=(z-1)(z+1) $$ has the only solutions $z=\pm 1$. This is true in the real numbers and stays true in the field of complex numbers.

Answer 5

From the fundamental theorem of algebra, you know that the equation $z^2=1$, a quadratic one, has at most two roots.

One of the roots is the obvious $z=1$, and with a few seconds of thinking, $z=-1$ is the other.

Answer 6

By far, the simplest solution is simply to take the square root of both sides:

$c+di = \pm 1$

which allows us to immediately equate real and imaginary parts to get $c = \pm 1, d = 0$.

Much simpler than squaring the LHS.