Find the multiplicative inverse of 2−3i2 - 3i2−3i.
Let z=2−3iz = 2 - 3iz=2−3i. Formula: z−1=z‾∣z∣2z^{-1} = \frac{\overline{z}}{|z|^2}z−1=∣z∣2z. z‾=2+3i\overline{z} = 2 + 3iz=2+3i. ∣z∣2=22+(−3)2=4+9=13|z|^2 = 2^2 + (-3)^2 = 4 + 9 = 13∣z∣2=22+(−3)2=4+9=13. z−1=2+3i13=213+i313z^{-1} = \frac{2 + 3i}{13} = \frac{2}{13} + i\frac{3}{13}z−1=132+3i=132+i133
If (x+iy)3=u+iv(x+iy)^3 = u + iv(x+iy)3=u+iv, show that ux+vy=4(x2−y2)\frac{u}{x} + \frac{v}{y} = 4(x^2 - y^2)xu+yv=4(x2−y2).
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