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Find the cube root of the decimal: 19.683.

The cube root of the decimal 19.683 is given by \(\begin{array}{l}\begin{array}{l} \sqrt[3]{19.683}\\=\sqrt[3]{\frac{19683}{1000}} \\ =\frac{\sqrt[3]{19683}}{\sqrt[3]{1000}} \\ =\frac{\sqrt[3]{3 \times 3 \times 3... View Article

If (z − 1) / (z + 1) is purely imaginary then what would be the locus of z?

Let z=x+iy \(\begin{array}{l}\frac{\mathrm{z}-1}{\mathrm{z}+1}=\frac{\mathrm{x}+\mathrm{iy}-1}{\mathrm{x}+\mathrm{i} \mathrm{y}+1}\\ =\frac{(\mathrm{x}-1)+\mathrm{i} y}{(\mathrm{x}+1)+\mathrm{i} y} \cdot \frac{(\mathrm{x}+1)-\mathrm{i} \mathrm{y}}{(\mathrm{x}+1)-\mathrm{i} \mathrm{y}}\\ =\frac{(\mathbf{x}-\mathbf{1})(\mathbf{x}+\mathbf{1})-\mathbf{i}^{2} \mathbf{y}^{2}+\mathrm{i} \mathbf{y}[\mathbf{x}+\mathbf{1}-(\mathbf{x}-1)]}{(\mathbf{x}+1)^{2}-\mathbf{i}^{2} \mathbf{y}^{2}} \frac{z-1}{z+1}\\ =\frac{x^{2}-1+y^{2}+i 2 y}{(x+1)^{2}+y^{2}}\\ \frac{\mathrm{z}-1}{\mathrm{z}+1}=\frac{\mathrm{x}^{2}-1+\mathrm{y}^{2}}{(\mathrm{x}+1)^{2}+\mathrm{y}^{2}}+\mathrm{i}... View Article