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- Homework Statement
- Let ##z=a+bi##, where ##a## and ##b## are real numbers. If Let ##z=a+bi##, where ##a## and ##b## are real numbers. If $$\frac {z}{z^*}=c+di$$, where ##c## and ##d## are real, prove that ##c^2+d^2=1##

- Relevant Equations
- Complex numbers

Easy questions, just a lot of computation...

$$\frac {z}{z^*}=\frac {a+bi}{a-bi} ×\frac {a+bi}{a+bi}$$

$$c+di=\frac {a^2-b^2}{a^2+b^2}+\frac {2abi}{a^2+b^2}$$

$$⇒c^2= \frac {a^4-2a^2b^2+b^4}{(a^2+b^2)^2}$$

$$⇒d^2= \frac {4a^2b^2}{(a^2+b^2)^2}$$

Therefore, $$c^2+d^2= \frac {a^4-2a^2b^2+b^4}{(a^2+b^2)^2}+\frac {4a^2b^2}{(a^2+b^2)^2}=1$$

A different approach would be appreciated...

$$\frac {z}{z^*}=\frac {a+bi}{a-bi} ×\frac {a+bi}{a+bi}$$

$$c+di=\frac {a^2-b^2}{a^2+b^2}+\frac {2abi}{a^2+b^2}$$

$$⇒c^2= \frac {a^4-2a^2b^2+b^4}{(a^2+b^2)^2}$$

$$⇒d^2= \frac {4a^2b^2}{(a^2+b^2)^2}$$

Therefore, $$c^2+d^2= \frac {a^4-2a^2b^2+b^4}{(a^2+b^2)^2}+\frac {4a^2b^2}{(a^2+b^2)^2}=1$$

A different approach would be appreciated...

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