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Question Number 168208 by MathsFan last updated on 06/Apr/22

If (x+yi)^4 =a+bi, show that a^2 +b^2 =(x^2 +y^2 )^4

$$\:{If}\:\left({x}+{y}\boldsymbol{{i}}\right)^{\mathrm{4}} ={a}+{b}\boldsymbol{{i}}, \\ $$$$\:{show}\:{that}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{4}} \\ $$

Commented by MJS_new last updated on 06/Apr/22

you can do it yourself (1) expand (x+yi)^4 =term1 (2) a=real (term1) ∧ b=imag (term1) (3) expand a^2 +b^2 =term2 (4) sort term2 and you should see the factors 1 4 6 4 1 ⇒ there you are

$$\mathrm{you}\:\mathrm{can}\:\mathrm{do}\:\mathrm{it}\:\mathrm{yourself} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{expand}\:\left({x}+{y}\mathrm{i}\right)^{\mathrm{4}} =\mathrm{term1} \\ $$$$\left(\mathrm{2}\right)\:{a}=\mathrm{real}\:\left(\mathrm{term1}\right)\:\wedge\:{b}=\mathrm{imag}\:\left(\mathrm{term1}\right) \\ $$$$\left(\mathrm{3}\right)\:\mathrm{expand}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{term2} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{sort}\:\mathrm{term2}\:\mathrm{and}\:\mathrm{you}\:\mathrm{should}\:\mathrm{see}\:\mathrm{the} \\ $$$$\:\:\:\:\:\mathrm{factors}\:\mathrm{1}\:\mathrm{4}\:\mathrm{6}\:\mathrm{4}\:\mathrm{1}\:\Rightarrow\:\mathrm{there}\:\mathrm{you}\:\mathrm{are} \\ $$

Answered by mr W last updated on 06/Apr/22

generally we have if z_1 =z_2 , then ∣z_1 ∣=∣z_2 ∣. if z_1 =z_2 ^n , then ∣z_1 ∣=∣z_2 ∣^n . applying this we get from (x+yi)^4 =a+bi ∣x+yi∣^4 =∣a+bi∣, or ((√(x^2 +y^2 )))^4 =(√(a^2 +b^2 )), or (x^2 +y^2 )^4 =a^2 +b^2

$${generally}\:{we}\:{have} \\ $$$${if}\:{z}_{\mathrm{1}} ={z}_{\mathrm{2}} ,\:{then}\:\mid{z}_{\mathrm{1}} \mid=\mid{z}_{\mathrm{2}} \mid. \\ $$$${if}\:{z}_{\mathrm{1}} ={z}_{\mathrm{2}} ^{{n}} ,\:{then}\:\mid{z}_{\mathrm{1}} \mid=\mid{z}_{\mathrm{2}} \mid^{{n}} . \\ $$$${applying}\:{this}\:{we}\:{get}\: \\ $$$${from}\:\left({x}+{yi}\right)^{\mathrm{4}} ={a}+{bi} \\ $$$$\mid{x}+{yi}\mid^{\mathrm{4}} =\mid{a}+{bi}\mid,\:{or} \\ $$$$\left(\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right)^{\mathrm{4}} =\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} },\:{or} \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{4}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$

Commented by MathsFan last updated on 19/Apr/22

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Answered by Mathspace last updated on 06/Apr/22

(x+yi)^4 =a+bi ⇒(x−yi)^4 =a−bi ⇒ (a+bi)(a−bi)=(x+yi)^4 (x−yi)^4 ⇒ a^2 +b^2 =((x+yi)(x−yi))^4 ⇒ a^2 +b^2 =(x^2 +y^2 )^4

$$\left({x}+{yi}\right)^{\mathrm{4}} ={a}+{bi}\:\Rightarrow\left({x}−{yi}\right)^{\mathrm{4}} ={a}−{bi}\:\Rightarrow \\ $$$$\left({a}+{bi}\right)\left({a}−{bi}\right)=\left({x}+{yi}\right)^{\mathrm{4}} \left({x}−{yi}\right)^{\mathrm{4}} \:\Rightarrow \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\left(\left({x}+{yi}\right)\left({x}−{yi}\right)\right)^{\mathrm{4}} \:\Rightarrow \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{4}} \\ $$

Commented by mr W last updated on 07/Apr/22

how do you get (x+yi)^4 =a+bi ⇒(x−yi)^4 =a−bi? it is not obvious.

$${how}\:{do}\:{you}\:{get} \\ $$$$\left({x}+{yi}\right)^{\mathrm{4}} ={a}+{bi}\:\Rightarrow\left({x}−{yi}\right)^{\mathrm{4}} ={a}−{bi}? \\ $$$${it}\:{is}\:{not}\:{obvious}. \\ $$

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