Collect the real and imaginary parts separately. Subtracting a negative contribution adds it.

The four points are \((2,1),(-2,-1),(2,-1),(-2,1)\). Their sides lie on two horizontal and two vertical lines. Adjacent sides are perpendicular, so they form a rectangle with center at the origin.

A sum of real squares is zero exactly when both are zero. Thus \(x=y=0\), precisely \(z=0\).

The point is in the second quadrant.

If either number is zero, both sides are zero. Otherwise write both in polar form. The product has modulus equal to the product of their positive radii, because its exponential factor lies on the unit circle.

Squaring either candidate returns the original number.

If \(v^2=u^2\), factor to obtain \((v-u)(v+u)=0\). Complex numbers have no zero divisors, so \(v=u\) or \(v=-u\). Since \(u\ne0\), these roots are distinct.

The modulus is \(2\), and the arguments are \(0,2\pi/3,4\pi/3\). The roots are \(2,-1+i\sqrt3,-1-i\sqrt3\), whose sum is zero. More generally let \(\zeta=e^{2\pi i/n}\) for n≥2. The geometric sum satisfies \((1-\zeta)(1+\zeta+\cdots+\zeta^{n-1})=1-\zeta^n=0\). Since \(\zeta\ne1\), the sum is zero, and multiplying by any initial root preserves that conclusion.

The expression requires \(z\ne i\). Squaring gives \(x^2+y^2=4[x^2+(y-1)^2]\). Rearranging and completing the square:

This circle has center \(4i/3\) and radius \(2/3\). The excluded point \(i\) does not satisfy the circle equation, so no point of this circle must be removed.

For \(z\ne0\), the real part is \(x/(x^2+y^2)\). Thus \(x=x^2+y^2\).

Remove the origin: it satisfies the final polynomial equation but the original reciprocal is undefined there.

Since \(\operatorname{Re}(z\bar w)\le|z\bar w|=|z||w|\),

Both sides are nonnegative, so square roots preserve the inequality. Equality holds when a number is zero or the two nonzero numbers point in the same direction.

Approach. Combine real parts and imaginary parts separately.

• First. \(2+1+(3-6)i=3-3i\)
• Second. \(-8+20i+7+2i=-1+22i\)

Approach. Use coordinate addition, then the parallelogram rule.

For \(z=x+iy\) and \(w=s+it\), real addition gives \(z+w=(x+s)+i(y+t)=(s+x)+i(t+y)=w+z\). Geometrically, the two possible orders traverse adjacent sides of the same parallelogram and reach its opposite vertex. This also covers collinear vectors, where the parallelogram is degenerate.

Approach. Conjugation changes the sign of the imaginary coordinate.

The points are \((2,1),(-2,-1),(2,-1),(-2,1)\). In cyclic order they form a rectangle centered at the origin, with horizontal sides of length \(4\) and vertical sides of length \(2\). Both coordinate axes are symmetry axes.

Approach. Interpret the modulus as distance from the origin.

By the distance formula, \(|2+i|=\sqrt{2^2+1^2}=\sqrt5\).

Approach. Expand and replace every occurrence of the square of i by −1.

The first is the defining rule for \(i\). Expanding the others gives \(1+2i+i^2=2i\) and \(4-4i+5i-5i^2=9+i\).

Approach. Identify the quadrant before choosing an argument.

Both moduli equal \(\sqrt2\). Their arguments are respectively \(\pi/4\) and \(-\pi/4\) modulo \(2\pi\). Thus \(1+i=\sqrt2e^{i\pi/4}\) and \(1-i=\sqrt2e^{-i\pi/4}\).

Approach. Distribute first; collect real and imaginary terms last.

• a. \(4+8i-10+2i=-6+10i\)
• b. \(1+\pi\sqrt2+2\sqrt2i\)
• c. \(8+2i-1-6i=7-4i\)
• d. \(1-2(3-11i)=-5+22i\)

Approach. For a product of three factors, simplify the first two before multiplying the third.

• a. \(1-i^2=2\)
• b. \(-10+15i-20i+30i^2=-40-5i\)
• c. \(2i(1+i)=-2+2i\)
• d. \((11+2i)(5+6i)=55+76i+12i^2=43+76i\)

Approach. Prove the conjugation identity before using it in the modulus-of-a-product proof.

Write \(z=x+iy\) and \(w=s+it\).

• a. \(|kz|=\sqrt{k^2x^2+k^2y^2}=|k|\sqrt{x^2+y^2}=|k||z|\)
• b. \(z\bar z=(x+iy)(x-iy)=x^2+y^2=|z|^2\)
• d. \(\overline{zw}=(xs-yt)-i(xt+ys)=(x-iy)(s-it)=\bar z\bar w\)
• c. Using (b) and (d), \(|zw|^2=zw\overline{zw}=z\bar z w\bar w=|z|^2|w|^2\). Both sides of the desired identity are nonnegative, so taking their nonnegative square roots proves it.

Approach. Use real distributivity on the real and imaginary coordinates.

Let \(z=x+iy,w=s+it,u=a+ib\). The left side is \(x(s+a)-y(t+b)+i[x(t+b)+y(s+a)]\). Distributivity of real multiplication rewrites this as \((xs-yt)+i(xt+ys)+(xa-yb)+i(xb+ya)\), which is precisely \(zw+zu\).

Approach. Separate the real and imaginary parts of the exponent.

Euler’s definition gives \(e^{1+i\pi}=e(\cos\pi+i\sin\pi)=-e\) and \(e^{i\pi/2}=i\). In general \(e^{a+ib}=e^a\cos b+i e^a\sin b\).

Approach. Use the real exponential law and the two trigonometric addition formulas.

Set \(z=x+iy,w=s+it\). By definition and the real exponential law,

The real angle-addition identities turn the bracket into \(\cos(y+t)+i\sin(y+t)\). This is exactly the defining expression for \(e^{(x+s)+i(y+t)}=e^{z+w}\).

Approach. Separate even and odd powers; justify the rearrangement by absolute convergence.

For real \(t\), the exponential series at \(it\) is absolutely convergent, so it may be separated into its even and odd terms:

The two resulting series are precisely the real cosine and sine series.

Approach. Evaluate sine and cosine at zero.

The exponent \(i0\) is zero, so the ordinary rule gives \(e^0=1\). Euler’s formula independently gives \(\cos0+i\sin0=1+i0=1\). These two descriptions agree.

Approach. Use modulus and the signs of both coordinates to choose the correct angle.

• First. Modulus \(2\), argument \(2\pi/3\): \(2e^{2\pi i/3}\).
• Second. Modulus \(4\), argument \(\pi/2\): \(4e^{i\pi/2}\).
• Third. Modulus \(10\), argument \(-\pi/3\): \(10e^{-i\pi/3}\).

Approach. Multiply numerator and denominator by the denominator’s conjugate.

• a. \(1/i=-i\)
• b. \(\frac1{1+i}=\frac{1-i}{2}\)
• c. \((1+i)(-i)=1-i\)
• d. \(\frac{(4+i)(1+2i)}5=\frac25+\frac95i\)
• e. \(i/4=0+(1/4)i\)

Approach. Invert the modulus and negate the argument.

If \(z=re^{i\theta}\) with \(r>0\), put \(w=r^{-1}e^{-i\theta}\). The multiplication rule gives \(zw=1\). An inverse is unique because \(zu=zv=1\) implies \(u=u(zv)=(uz)v=v\). Zero has no inverse, since \(0w=0\) for every \(w\).

Approach. Squaring doubles the angle and squares the modulus.

Let \(z=re^{i\theta}\) with \(r>0\). If \(w=\rho e^{i\phi}\) and \(w^2=z\), then \(\rho^2=r\) and \(2\phi=\theta\pmod{2\pi}\). Consequently \(\rho=\sqrt r\) and \(\phi=\theta/2\pmod\pi\). The two distinct roots are \(\sqrt r e^{i\theta/2}\) and its negative. Their squares equal \(z\); the preceding necessity proves there are no others. Zero has only the root zero.

Approach. Check the real part and twice the product of the proposed root’s coordinates.

Write \(r=\sqrt{x^2+y^2}\) and \(a=\sqrt{(r+x)/2},\ b=\sqrt{(r-x)/2}\). These are real and nonnegative because \(r\ge|x|\). Then \(a^2-b^2=x\) and \(2ab=\sqrt{r^2-x^2}=|y|\). Hence for \(y\ge0\), \((a+ib)^2=x+iy\); the other root is \(-a-ib\). For \(y<0\), use \(a-ib\) and its negative. At \(z=0\) the two displayed values coincide, so there is only one root.

Approach. Use half-arguments for simple polar forms and the Cartesian formula otherwise.

• a. \(\pm i\)
• b. \(\pm2i\)
• c. \(\pm(1+i)/\sqrt2\)
• d. \(\pm(1+i)\)
• e. \(\pm(1-i)/\sqrt2\)
• f. \(\pm(1-i\sqrt3)/\sqrt2\)
• g. \(\pm(\sqrt{\sqrt2+1}+i\sqrt{\sqrt2-1})\)
• h. \(\pm2\)

For (f), squaring gives \((1-2i\sqrt3-3)/2=-1-i\sqrt3\). For (g), the difference of squared real and imaginary parts is \(2\) and twice their product is \(2\sqrt{(\sqrt2+1)(\sqrt2-1)}=2\). The other cases follow by direct squaring. Exercise 12 proves each nonzero number has no further square roots.

Approach. Complete the square without assuming the coefficients are real.

Multiplying by \(4a\) and completing the square gives the equivalent equation \((2az+b)^2=b^2-4ac\). Every complex number has a square root, so

If the discriminant is zero these coincide; otherwise the two square roots are distinct, and division by \(2a\ne0\) preserves distinctness. Completing the square was reversible, so these are all roots.

Approach. Expand the modulus squared as a number times its conjugate.

Conjugation gives \(\overline{z-\bar k}=\bar z-k\). Therefore

Subtract \(|k|^2\) to obtain the required expression. Note that the center is \(\bar k\), not \(k\).

Approach. Translate the vertex to zero before using polar form.

Translation by \(-q\) places the vertex at zero and changes the two rays to vectors \(p-q\) and \(r-q\). Write these as \(\rho e^{i\alpha}\) and \(\sigma e^{i\beta}\) with positive moduli. Their quotient is \((\sigma/\rho)e^{i(\beta-\alpha)}\), whose argument is precisely the counterclockwise directed angle from the first ray to the second, modulo \(2\pi\). For the ordinary undirected angle, choose the representative in \([0,\pi]\) by taking the smaller angular separation.

Approach. Factor a constant out of each modulus, then use distance from a center.

• a–d. Circles with (center, radius) respectively \((0,1),(0,2),(1,2),(-i,3)\).
• e. Since \(|iz+1|=|i(z-i)|=|z-i|\), the locus is the circle centered at \(i\) with radius \(6\).
• f. Divide by \(4\): \(|z+i/2|=1/8\). The center is \(-i/2\) and radius \(1/8\).
• g. The strict exterior \(|z-i/3|>2\); omit the boundary.
• h. The open disk \(|z-(-10+10i)|<50\).
• i. Squaring gives \((x-1)^2+y^2=x^2+y^2\), hence the perpendicular bisector \(x=1/2\).
• j. Squaring gives \(x^2+(y+1)^2=x^2+(y-1)^2\), hence \(y=0\).
• k. The vertical line \(x=1\).
• l. The open half-plane above \(y=-1\).

Approach. Rationalize real/imaginary parts, or square a modulus equation. Never restore a pole after simplifying.

• a. For \(z=x+iy\ne0\), \(\operatorname{Re}(1/z)=x/(x^2+y^2)\). Thus \((x-1/2)^2+y^2=1/4\), with \(0\) removed.
• b. \(\operatorname{Im}(1/z)=-y/(x^2+y^2)\), so \(x^2+(y+1)^2=1\), again with \(0\) removed.
• c. Replace \(x\) by \(x-1\) in (a): \((x-3/2)^2+y^2=1/4\), with \(z=1\) removed.
• d. Since \(z/(z-i)=1+i/(z-i)\), its imaginary part is \(x/[x^2+(y-1)^2]\). Multiplying by the positive denominator gives \((x-1)^2+(y-1)^2>1\). The pole \(i\) lies on the excluded boundary.
• e. \(|z-1|=1\): center \(1\), radius \(1\); the pole is not on this circle.
• f. Squaring gives \(15x^2+15y^2+32y+16=0\), or \(x^2+(y+16/15)^2=16/225\). Center \(-16i/15\), radius \(4/15\); the pole \(-i\) is not on it.
• g. Squaring gives \(4(x^2+y^2)=(x+2)^2+y^2\), so \((x-2/3)^2+y^2=16/9\). The pole \(-2\) is not on this circle.
• h. Since \(|iz-2|^2=x^2+(y+2)^2\), squaring yields \(8x^2+8y^2-2x+36y+35=0\). Thus \((x-1/8)^2+(y+9/4)^2=45/64\); center \(1/8-9i/4\), radius \(3\sqrt5/8\). The pole \(-2i\) is not on it.

Approach. Expand only the imaginary part.

Put \(\alpha=a+ib,\ \beta=c+id,\ z=x+iy\). The equation becomes \(bx+ay+d=0\). This is a line exactly when \(\alpha\ne0\). Conversely, for a line \(Ax+By+C=0\) with \((A,B)\ne(0,0)\), choose \(\alpha=B+iA,\ \beta=iC\). If \(\alpha=0\), the locus is the whole plane when \(d=0\) and empty otherwise.

Approach. Expand a difference of squares, then differentiate the two exponentials.

The real chain rule gives \((\cosh t)^{\prime}=(e^t-e^{-t})/2=\sinh t\) and \((\sinh t)^{\prime}=(e^t+e^{-t})/2=\cosh t\).

Approach. Recover the parameter from x+y.

Exercise 21 proves the equation, and \(\cosh t>0\). Conversely, on this branch \(x=\sqrt{1+y^2}>|y|\), so \(x+y>0\). Define \(t=\ln(x+y)\). Because \((x+y)(x-y)=1\), we have \(e^t=x+y\) and \(e^{-t}=x-y\). Taking their half-sum and half-difference recovers \(x=\cosh t,y=\sinh t\). The formula for \(t\) is unique.

Approach. Compare the equations of the unit circle and the unit hyperbola.

The counterparts are \(\cos^2t+\sin^2t=1\), \((\sin t)^{\prime}=\cos t\), and \((\cos t)^{\prime}=-\sin t\). The point \((\cos t,\sin t)\) runs around the unit circle, whereas \((\cosh t,\sinh t)\) runs along a hyperbola branch. Circular functions are periodic; the hyperbolic parametrization is one-to-one on the real line. The changed signs are important, so the analogies are not literal equalities of all properties.

Writing \(z=x+iy\) gives \((1-t^2)(x^2+y^2)+2t^2x-t^2=0\). At \(t=1\), this is \(x=1/2\), the perpendicular bisector of zero and one. Otherwise completing the square gives center \(t^2/(t^2-1)\) on the real axis and radius \(t/|1-t^2|\). Squaring introduced no extra points because both original sides are nonnegative. The excluded parameter is not a missing locus: it produces a line instead of a circle.

The two expansions are \(|z|^2+|w|^2+z\bar w+\bar z w\) and \(|z|^2+|w|^2-z\bar w-\bar z w\). Adding cancels the mixed terms and proves the identity, including zero or collinear inputs. In a parallelogram with adjacent side vectors z,w, the diagonal vectors are z+w and z-w. Thus the sum of squared diagonal lengths equals the sum of squared lengths of all four sides.